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ELEC 103 UNIT 2
Unit two is essentially made up of three individual areas that all utilize the information from the prior
unit. The three major areas of this unit, along with the subdivisions, are as follows:
1. Scientific Notation
2. Engineering Notation (Shorthand Notation)
3. Ohms Law & Power Calculations in:
a. Series Circuits
b. Parallel Circuits
c. SeriesParallel Circuits
Scientific notation is a method of representing numbers in a form that makes it easier to enter the
number into a calculator and also easier to utilize the number in normal calculations. In the electronics
industry, numbers are used that are very large, as in the case of resistors, or very small, as in the case of
the values for capacitors. Since most calculators have only a seven or eight number display, it is very
difficult to enter a number like 0.000000000012. It simply does not fit. However, this is the value of a
capacitor that would normally be referred to as a 12 pico Farad capacitor in normal conversation, but
would be entered into a calculator as 1.2 x10-11. The number 1.2 x10-11 is exactly equal to the number
0.000000000012 but is simply presented in another form. The symbols after the number 1.2, are
written in scientific notation. You must learn how to use scientific notation if you are going to use the
calculator for scientific calculations. You must be able to perform scientific calculations in this field of
endeavor.
Learning SCIENTIFIC NOTATION means learning how to write a number in a form that can be
used for calculator entries and also converted into ENGINEERING NOTATION (shorthand
notation) for conversational and writing usage. An example of converting a number into scientific
notation and then converting the scientific notation into engineering notation follows: The number
0.000000000012 Farads can be written as 1.2 x10-11 Farads and can also be written as 12 pico Farads.
They all mean exactly the same thing. The regular number format is difficult to verbalize or use in
normal conversation and almost impossible to use on a calculator. However, the scientific notation
format can be used on a calculator very nicely. Of Course, the "shorthand" notation can be verbalized
and put into print much more readily than the other two formats. There is a need for all three formats in
our field and therefore, you must learn all three formats.
NUMBER FORMAT TO SCIENTIFIC NOTATION CONVERSION
The rules for converting a number into scientific notation are simple if you follow the few steps
necessary to perform the conversion.
1.
Move the decimal point from where it is presently located to a position located to the right
of the first integer. In the number used in the previous example, the decimal would be moved
from where it is located to a position between the one and the two. You would write the number
as 1.2, which places the decimal to the right of the first integer.
2.
When the number is written it should be followed by x10. The x10 is going to be referred to
as the carrier. The carrier is only in place to carry the important information. Therefore, the
number should presently be written as 1.2 x10.
3.
Count the number of places the decimal point was moved and write that number as a
superscript of the carrier. The converted number should now be written as 1.2 x1011 since the
decimal point was moved eleven places from where it was located to the present position after
the first integer.
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4.
The last part of the conversion procedure requires you to decide which way you would need to
move the decimal to get it back to where it was located before you moved it to the present
location. If you need to move the decimal point to the left to return it to the original
location, you must place a minus sign in front of the superscript number. That is the
situation with the number being presently used in this example. Therefore, in this example, the
number 0.000000000012 Farads when converted into scientific notation is 1.2 x10-11 Farads.
5.
If you need to move the decimal point to the right to return it to the original location, you
must place a plus sign in front of the superscript number. When this condition exists, the
plus sign can be placed in position or not placed in position. The sign is understood to be PLUS
when no sign is present and that is the preferred method. To put this procedure into practice
will require you to go through a few examples and then try to do a few problems on your own.
There is a resistor that has a value of two million four hundred thousand Ohms that needs to be
converted into scientific notation. The number is 2,400,000 Ohms. Move the decimal to the right of the
first integer. Alternatively, if you want to say it differently, say, move the decimal between the first and
second numbers. You would write 2.4 followed by the carrier. This would make the number 2.4 x10
with the superscript number and sign still to be added. Count the number of places the decimal point
was moved and use that number in the superscript position. The conversion presently is 2.4 x106 with
the sign in front of the superscript still to be added. It should be obvious that the decimal being located
between the two and the four, would have to be moved to the right to make the number the same as it
was originally. Therefore, a plus sign should be placed in front of the superscript number. The
conversion would make 2,400,000 Ohms equal 2.4 x10+6 Ohms, or 2.4 x106 Ohms, but the exponent
without the sign is preferred. Since events happen very quickly in electronics, we often discuss events
as happening in small parts of a second. An event happened in 0.00008 seconds and we need to convert
that time into scientific notation to make some calculations.
The first step is to move the decimal point from the present location to a new location to the right of
the first integer. This would require moving the decimal point to a position to the right of the number
eight. Therefore, you would write the number followed by the carrier, which is 8 x10.
The second step is to count the number of places the decimal has been moved and use that number as
the superscript number for the carrier. Therefore, you would write 8 x105 since the decimal was moved
five places.
The third step is to determine in which direction you must move the decimal to return the number to
its original format. If you must move the decimal to the left to return the number to its original format,
a minus sign must be placed in front of the superscript number. On the other hand, if you must move
the decimal to the right to return the number to its original format a plus sign must be placed in front of
the superscript number.
In this example, you should have determined the direction needed to return the number to its original
format was to the left and therefore, the completed conversion is 8 x10-5 seconds. As you probably
know, our ears can hear frequencies between 20 Hertz (cycles per second) and 20,000 Hertz (Hz). We
need to use 19,500 Hz in scientific notation format, so therefore, must perform a conversion. The first
step in the conversion is to move the decimal point from its present location after the last zero to a
position between the first and second integer. That means the number, converted thus far, would be
1.95 x10. The second step in the conversion is to count the number of places the decimal point was
moved. You should count a movement of four places. This means the number four can be added as the
superscript of the carrier. That means the number, converted thus far, would be 1.95 x 104. The third
and last step is to determine which direction the decimal point must be moved to return the number to
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ELEC 103 UNIT 2
its original position. It should be obvious that the decimal must be moved to the right to get it back to
its original position. The final conversion of 19,500 Hz to scientific notation is such that the number is
1.95 x 10+4 or, as preferred, it should be written without the plus sign as 1.95 x 104 Hz.
1. 0.005 Amperes
In the figure on the right you see a listing of several numbers. Before going
2. 0.0000021 Sec
any further, you should make an effort to convert each one of the numbers in
3. 27000 Ohms
Figure 21 into scientific notation. The answers, in scientific notation
4. 5400000 Hz
format, for each one of the four numbers in the figure will be available in
FIGURE 21
Figure 22 on the next page. Keep in mind the three step procedure involved
in the conversion and follow it very carefully for smooth conversions.
SCIENTIFIC NOTATION TO NUMBER FORMAT CONVERSION
Next on the agenda is the conversion from scientific notation back into the original number format.
This conversion is the opposite of the conversion routine you have been learning using the numbers in
Figure 21 and Figure 22. Most importantly, you should learn how to convert regular numbers into
scientific notation. The conversion from scientific notation into number format is easy if you know
how the number got into scientific notation in the first place. Suppose the number 8 x103 Amperes is
the number, in scientific notation, that is going to be converted back into its original number format.
The first step in the conversion is to look at the superscript attached to the carrier. The superscript
number tells you how many places to move the decimal, and the sign indicates which way to move the
decimal point. Therefore, in the present example, 8 x10-3 Amperes, the decimal should be moved three
places. The second step is to determine which way the decimal is to be moved. Since the sign is
minus, the decimal should be moved three places to the left from its present position. The third step is
to move the decimal to the new position and attach the correct units to the number. Therefore, 8 x10-3
Amperes converts to 0.008 Amperes. It is obvious that once you know how to convert into scientific
notation it is quite easy to convert back to number format.
The first step in the conversion is to determine the number of places to move the decimal point to
return the decimal to its original position. You can determine that number by reading the superscript
number attached to the carrier. In this example, the number of places to move the decimal is twelve
places. The second step is to determine the direction to move the decimal. If the sign of the superscript
number is minus, the decimal will be moved to the left. If the sign of the superscript number is plus, or
there is no sign, the decimal will be moved to the right. In this example, the decimal will be moved to
the left. The third step involves moving the decimal the appropriate number of places in the proper
direction and attaching the unit designator to the number. Of course, the carrier and the carrier
superscript are discarded. The number for this example is 9.1 x10-12 Farads, which converts to
0.0000000000091 Farads.
In Figure 23, you will find five numbers in scientific
Scientific Notation Number Format
notation. Perform a scientific notation format to number
1. 3.9 x104 Ohms
format conversion on each one of the numbers in the figure.
-9
The answers are on the next page in Figure 24. WAIT !
2. 5 x10 Farads
Do not look until you have worked out the answers, then
3. 9.050 x107 Hz
look.
4. 5.4 x10-6 Amps
5. 1.5 x102 Volts
FIGURE 23
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ELEC 103 UNIT 2
The answers to the two previous questions.
Number Format
Scientific Notation
Scientific Notation
Number Format
1. 3.9 x104 Ohms
39,000 Ohms
2. 5 x10-9 Farads
0.000000005 Farads
3. 9.050 x107 Hz
90,500,000 Hz
0.005Amps
=
5 x 10-3 Amps
0.0000021 Sec
=
2.1 x10-6 Sec
27000 Ohms
=
2.7 x104 Ohms
4. 5.4 x10-6 Amps
0.0000054 Amps
54000000 Hz
=
5.4 x107 Hz
5. 1.5 x102 Volts
150 Volts
FIGURE 24
FIGURE 22
ENGINEERING NOTATION
Engineering notation requires a letter or a word be substituted in place of the scientific notation carrier.
In other words, if the number under the scientific notation column, in problem four of Figure 24 was
going to be used in a written report it would not be written as you see it, 5.4 x10 -6 Amps. The carrier
and its superscript would be thrown away and replaced with a letter or a word that is known to be equal
to that particular carrier. The written version would be 5.4µA or could also be written as 5.4
microAmps. There are standard symbols and words that are the equivalent of specific scientific
notation format. These symbols and words make up the shorthand notation that can be used to replace
scientific notation and number format for the written or spoken word. These symbols, shorthand words
and their equivalent value in scientific notation are called engineering notation.
One of the first things to look at in Figure 25 is the
PREFIX
SYMBOL POWER OF 10
scientific notation equivalent column. Note that all the
Tera
T
x1012
exponents are in three unit increments. Specifically
running from plus 12 to minus 12. This introduces a
Giga
G
x109
problem when you want to convert a number from
Mega
M
x106
scientific notation format into engineering format.
Kilo
k
x103
Suppose you had a number similar to number one in
Figure 24 and you wanted to convert it into
Units
x100
engineering notation. The exponent of the carrier is plus
milli
m
x10-3
four and there is only a plus three or plus six listed for
engineering notation conversion. The solution is to put
micro
x10-6

4
the number 3.9 x10  into engineering notation format
nano
n
x10-9
with a carrier similar to one of those in Figure 25
under power of 10.
pico
p
x10-12
You should use one of those listed that will allow you to
FIGURE 25
have a whole number followed by one of engineering
notation numbers that appear in the table.
The first step is to convert the scientific notation format back to number format. To do that, the
decimal point should be moved four places to the right and the number becomes 39,000.
The second step is to move the decimal point either three places, six places, or nine places etc., so the
number and the carrier exponent agree with one of the listings in the engineering format table. If the
decimal is moved three places, the number becomes 39 x103. If the decimal is moved six places, the
number becomes 0.039 x106. The one that is acceptable is 39 x103 since it is a whole number and
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ELEC 103 UNIT 2
not a decimal number. Now that the number has been converted into engineering notation format with
a ± 3, ± 6, ± 9, or ± 12 superscript value, the shorthand substitution can be made. The number 39
x103 can be made into 39k by substituting k for x103.
Look at the second number in Figure 24. Convert that number into engineering notation (shorthand
format) using two possible formats, and in addition, show two versions of each format. To put the
second number in Figure 24 into engineering format is easy, since x10-9 is listed in the engineering
format table. The number 5 x10-9 Farads could be written as 5 nano Farads and could also be written as
5 nF. The second conversion will be more difficult. Just remember that the number, when converted,
should be a whole number and not a decimal number.
To convert the scientific notation format into engineering format, move the decimal in increments of
three until you have a whole number. Since, in the previous paragraph, the decimal was moved until
the number of decimal places moved was nine, the decimal should be moved an additional three places
in the same direction. This makes the number 5000, and therefore, the decimal would have been moved
a total of twelve places. Therefore, the number should be written, in engineering notation, as 5000 x1012
Farads. Of course, this is exactly the same as 5 x10-9 Farads but in another form. The number, 5000
x10-12 Farads, can be converted to 5000 pF and can also be converted to 5000 pico Farads. All three
numbers in the last sentence are equal. In fact, 5 nF, 5 nano Farads, 5000 pF, and 5000 pico Farads are
all equal but just appear in different forms.
When a number appears with the decimal place after the first number and is followed by the carrier and
exponent, it is said to be in scientific notation. However, if one wants to use engineering notation, the
decimal does not need to be located after the first number. The primary goal is to have the carrier
exponent be made into ± 3, ± 6, ± 9, or ± 12. When this occurs, the word substitution can be made for
the carrier and its exponent. The requirement that the decimal must be placed to the right of the
first integer in scientific notation, is not a requirement in engineering notation.
Engineering notation requires that the main number be a whole number and the carrier exponent be ± 3,
± 6, ± 9, or ± 12 followed by the appropriate unit designator. Unlike scientific notation, the main
number is not required to be a single integer. The main number can be any number of integers.
However, it must be followed by the appropriate carrier exponent so the equivalent shorthand notation
can be substituted for the carrier and its exponent.
For Figure 22, convert the numbers in the scientific notation column into engineering notation.
The answers will be found in Figure 26 on the next page.
This unit, so far, has covered the conversion of ordinary numbers into scientific notation. The
conversion of scientific notation into ordinary numbers. The conversion of ordinary numbers into
engineering notation. In addition, of course, if you can convert from scientific notation into ordinary
numbers, you can convert from engineering notation into ordinary numbers. The answers to the
conversions from scientific notation into engineering notation appear in word format and in symbol
format.
OHMS LAW
Ohms Law was discovered in 1827 by Georg Simon Ohm. Ohms Law applies to an entire circuit or to
any part of a circuit. Thus, the potential difference or voltage drop across any part of the circuit or
conductor equals the current (I) flowing in the conductor multiplied by the resistance (R) of that
conductor, or E = IxR.
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ELEC 103 UNIT 2
Scientific Notation
Engineering Notation
-3
1. 5 x10 Amps =
5 milliAmps or 5 mA
2. 2.1 x10-6 Sec =
2.1 microSec or 2.1µs
3. 2.7 x104 Ohms =
27 kiloOhms or 27 k
4. 5.4 x107 Hz =
54 MegaHertz or 54 MHz
On the left is the answer to the question from
the previous page.
FIGURE 26
Ohms Law applies to all linear conductors. Ohms Law does not apply to nonlinear conductors and
devices such as ionized gases, semiconductors, and other devices when operated in the nonlinear part
of their characteristics.
Figure 27 shows Ohms Law in three forms. The main form, I =
E=IxR
E  R, is the form used to describe Ohms Law. The Law states
that the current (I) is directly proportional to the voltage (E)
I=ER
and inversely proportional to the resistance (R). The current
is directly proportional to the voltage means that if the voltage is
R = E I
increased, the current in the circuit will increase in the same
FIGURE 27
proportion.
Figure 28 shows a mathematical analysis of the
Directly Current Inversely
statements in the last paragraph to help you better Current
To
understand Ohms Law. The current is directly Proportional To The Proportional
Voltage
The
Resistance
proportional to the voltage means that if the voltage is
I=ER
I = E R
doubled the current will double. This is shown in Figure
28 in the boxes numbered one and two. If you compare
1. 2A= 4V  2
4. 2A= 4V  2
the boxes that are numbered one and three, you see that
2. 4A= 8V  2
5. 1A= 4V  4
when the voltage is halved the current halves. This means
3. 1A= 2V  2
6. 4A= 4V  1
the current does just the same as the voltage does, and
FIGURE 28
does it in the same proportion. So when the voltage
doubles the current doubles and when the voltage halves the current halves. The boxes numbered four,
five and six, in Figure 28, show an inverse proportionality exists between the current and the
resistance. If the resistance is doubled, the current will halve and when the resistance is halved, the
current will double. This is a mathematical analysis of the verbal expression of Ohms Law, which
states the current is directly proportional to the voltage and inversely proportional to the resistance.
Figure 27 shows a circle with Ohms Law in mathematical format inside the divided circle. It is a way
to help you remember the three different versions of Ohms Law that are shown to the right of the
circle. In a problem, when you want to calculate the voltage, you place your finger over the E in the
divided circle, which represents the voltage. In addition, as you can see, the the R and I are next to each
other indicating multiplication as the uppermost formula also indicates. If you place a finger over the I
within the circle, you will see the E over the R as the middle formula indicates. The circle is a method
for you to use to remember the three versions of Ohms Law so you can calculate the voltage, the
current, and the resistance. However, after using Ohms Law for some problem solving, you will rely
less on the circle method to remember Ohms Law.
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ELEC 103 UNIT 2
OHMS LAW AND SERIES CIRCUITS
You have seen series circuits previously, in unit one, when you performed voltage and current
measurements. In this section of the present unit, you will be learning how to solve series circuit
problems for voltage, current, resistance and power. The definition of a series circuit is, the series
circuit provides only one path for the current to flow in the circuit, from the negative terminal of the
source to the positive terminal of the source through a resistive path. There are three rules that apply to
the series circuit. You must know the rules, in addition to Ohms Law, to solve series circuit problems.
The rules are related to voltage, current and resistance and can be written in word format and in
mathematical format.
1.
The rule for voltage states that the applied voltage or source voltage is equal to the sum of
the individual resistor voltage drops no matter how many resistors there are in the circuit.
The applied voltage (VA), the source voltage (VS), and the total voltage (VT) are all used
interchangeably. The mathematical expression would be written as VT = VR1 + VR2 + ••• +VRN
2.
The rule for current states that the current flowing in the circuit is the same everywhere in
the circuit. The total circuit current is the same as the current flowing in each resistor in the
circuit. Since there is only one path for current to flow it cannot be different in different parts of
the circuit. The mathematical expression is IT = IR1 = IR2 = ••• =IRN
3.
The rule for resistance states that the total resistance of the circuit is equal to the sum of the
individual resistances, no matter how many resistors there are in the circuit. The
mathematical expression for total circuit resistance is RT = R1 + R2 + ••• +RN
Figure 29 contains a series circuit that will be used to introduce you
to the techniques used to solve series circuit problems for voltage
drops, circuit current, and resistances. The method that is being
presented to solve series circuit problems is not the only method
available. If you wish to use another method, feel free to do so. In the
circuit of Figure 29, there are three GIVEN values. The voltage drop
across R1 is 6 Volts. The resistance of R2 is 3.9k and the total
current flowing in the circuit is 2mA.You are requested to FIND the
total or source voltage and the resistance of R1. To help remember the
values that have been GIVEN and the values that are to be calculated,
Figure 29
a tabulation format will be utilized.
Figure 210 illustrates the tabular format that will be
E
I
R
used for assistance in solving series circuit problems. As
R1
6V
*
you look at the table, notice there is a row for each
R2
3.9k
resistor in the circuit. There is also a row for the circuit
Total Ckt
*
2mA
totals. For each resistor there is a box for voltage, current
and resistance as there also is for total voltage, total
FIGURE 2-10
current and total resistance. The GIVEN quantities are
placed in their respective positions.
The values that are to be calculated are marked with an asterisk in the appropriate box. Most
importantly, Ohms Law can be used with values from one row only. You cannot pick a voltage from
one row, a resistance from another row and calculate a current. You must use only values on the same
row or horizontal line associated with the component in which you are interested. For instance, the
total circuit current must be used with the total circuit resistance to calculate the total circuit voltage.
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ELEC 103 UNIT 2
Additionally, the three rules for series circuits that were introduced earlier can be used with the tabular
format shown in Figure 210. For example, the voltage drop across R1 when added to the voltage drop
across R2 will equal the total circuit voltage. In addition, the resistance of R1 added to the resistance of
R2 will equal the total circuit resistance. In addition, since the current is the same everywhere in the
circuit, if you know one current you know all the currents. The current flowing through R1 is the same
as the current flowing through R2, which is the total circuit current.
The columns or vertical boxes are associated with the rules for voltage, for current, and for
resistance. The rows or horizontal boxes are associated with the Ohms Law formula. To start the
problem solving procedure, the table should be drawn as shown in Figure 210. Draw one row for each
resistor in the circuit then draw the last row for the total circuit values. There should be a column for
voltage, a column for current, and a column for resistance. Next, fill in the appropriate boxes with the
GIVEN information and finally, place an asterisk in the boxes where values are to be calculated. Once
all the preliminary work has been performed, the first step is to apply the three rules stated earlier, to
the table. The first rule, the one for voltage, will not benefit us. Since the total voltage is equal to the
sum of the individual resistor voltage drops, the formula requires only one unknown value. In this
example, only one voltage is known and two voltages are unknown. Therefore, the voltage rule cannot
help this time. The second rule, the rule for current can provide some help. The second rule states that
the current is the same everywhere in the circuit. Therefore, since the total circuit current is known, the
current through R2 and the current through R1 will be the same. Look at Figure 211, which shows the
appropriate boxes, filled in with the newly acquired information.
The third rule, the one for resistance, will not benefit us.
E
I
R
Since the total resistance is equal to the sum of the
R1
6V
2mA
*
individual resistances in the circuit, the formula requires
R2
2mA
3.9k
there be only one unknown value. In this example this is
not true, there are two unknowns.
Total Ckt
*
2mA
Now that the first step has been completed, the second
FIGURE 2-11
step can be started. The second step consists of looking
across the rows for a row with two known values and only one unknown value across the row. There
are two rows that have that qualification, the first row and the second row. You might say to yourself,
which row is the place to start. The answer to that question is that you should start where ever you see a
calculation possibility. If you can calculate the value for each box in the table, it makes sense that you
know everything about the circuit. Therefore, start where you see a calculation possibility and calculate
everything. When finished, you will have the answer to any question that might be asked about the
circuit. Looking at the first row in Figure 211, the value for resistance (R) is unknown but the values
for voltage (E) and current (I) are known. Therefore, using the circle method shown in Figure 27,
place your finger over R, the letter used for resistance. You should see that you need to divide the
voltage across R1 (ER1), by the current flowing through R1 (IR1), in order to calculate the resistance of
R1. The equation should look like this: (R1 = ER1  IR1). Substituting the proper values into the
equation, you should be dividing 6 Volts, the voltage across the resistor R1, by 2 mA, the current
flowing through R1. Therefore, you should see 6V divided by 0.002A, which equals 3000 Ohms. The
calculated resistance should be converted into engineering notation. When the calculated value has
been converted into engineering notation, the calculated value should be placed in the resistance box of
the R1 row of tabulated results for this problem.
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ELEC 103 UNIT 2
Figure 212 shows the results up to the present time.
E
I
R
REMEMBER, before starting the problem, and after each
R1
6V
2mA 3.0k
calculation you should apply the rules to each of the columns.
This time, the voltage rule is of no benefit and the current rule
R2
2mA 3.9k
is of no benefit. However, the resistance rule will allow us to
Total Ckt
*
2mA
calculate the total circuit resistance. The sum of the
FIGURE 2-12
individual resistances is equal to the total circuit resistance.
Therefore, the total resistance is equal to the sum of the 3k and the 3.9k resistors, which is 6.9k.
The lower right hand box, which represents the total circuit resistance, can be filled with 6.9k.
Figure 213 shows the results up to the present time.
E
I
R
Since the rules have just been applied, it is time to do
R1
6V
2mA
3.0k
another calculation. The second row and the last row
R2
2mA
3.9k
both qualify for a calculation since both rows have only
one unknown item. One of the FIND items was the total
Total Ckt
*
2mA
6.9k
voltage, so let us make the calculation using the data in
FIGURE 2-13
the last row. Refer to Figure 27. Place your finger over
the E, which represents voltage, which is the unknown in this calculation. You should see that the
current (I) should be multiplied by the resistance (R) in order to calculate the voltage (E). Therefore,
the equation should look like this: (ET = IT x RT). Substituting the numbers into the equation, you will
multiply 2mA by 6.9k, which means you will see 0.002A times 6900, which is equal to 13.8 Volts.
The calculated value should be placed in the total circuit voltage box of tabulated results for this
problem. This can be seen in Figure 214, which shows the results tabulated up to the present time,
including the last calculation.
The rules should be applied after each calculation.
E
I
R
Therefore, if you look at Figure 214, you should see
R1
6V
2mA
3.0k
that only one rule can be applied in this situation. The
voltage rule is the rule that can be applied since we know
R2
2mA
3.9k
two of the three voltages. The voltage rule states that the
Total Ckt
13.8V
2mA
6.9k
total circuit voltage equals the sum of the individual
resistor voltage drops. Since we know the total circuit
FIGURE 2-14
voltage and one of the resistor voltage drops, the
equation would look like this: (ER2 = ET  ER1). Subtracting 6 Volts from 13.8 Volts will give you the
voltage drop across R2. This calculated value should be placed in the appropriate box of tabulated
results for the problem. All the boxes are filled and therefore, the problem has been totally solved. You
can pick out the values you were requested to FIND, with no difficulty, since everything concerning the
circuit is known and is in one of the boxes in the tabulated format.
The final tabulated display for the original problem,
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stated in Figure 29 and just concluded, is presented in
R1
6V
2mA
3.0k
Figure 215. Look at it and review the features of using
R2
7.8V
2mA
3.9k
the tabulated format. The three rules are associated with
the columns. Look down the voltage column and note
Total Ckt
13.8V
2mA
6.9k
that 6 Volts plus 7.8 Volts equals 13.8 Volts. The current
FIGURE 2-15
column shows that the current is the same everywhere in
the circuit. Finally, the resistance column shows that the total resistance is equal to the sum of the
individual resistances. Also notice that the numbers on any row agree with any Ohms Law calculation
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ELEC 103 UNIT 2
you might perform using the data on that particular row. The tabulated format helps you remember the
rules, stores given and calculated values in a useful manner, and helps you use correct values for
calculations.
SERIES CIRCUITS AND POWER CALCULATIONS
When a current flows through a resistance there is work being done, and heat is being generated. An
example of this is the light bulb in you home. A 15W light bulb is not as bright as a 100W light bulb.
However, a 100W light bulb is considerably hotter than a 15W light bulb.
Both bulbs have current flowing through the resistance of the filament to
create the light that we utilize. However, since work is being done, heat is
being generated and is being dissipated into the atmosphere. Power (P) can
be calculated by multiplying the voltage (E) by the current (I) as shown in
Figure 216.
The unit of power is the WATT. Every resistive component that has a
Figure 216
current flowing through it, will have heat generated by the work that is
being performed. The component must get rid of the generated heat, dissipate the heat, so it does not
overheat and malfunction. To be able to calculate the power a component must dissipate, or a source
must supply to a circuit, you can use the circle format shown in Figure 216. This circle format works
exactly the same as the one you used for Ohms Law. The three basic forms of the power formula are
also displayed so you can verify the power circle method. Remember, the unit of power is the WATT
and power is calculated by multiplying voltage times current. Refer to Figure 215 and visualize
another column next to the resistance column. This column would have a P for power in the uppermost
box. The rule for the power is the total power being supplied to the circuit is equal to the sum of
the power being dissipated by the individual
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P
components in the circuit. The additional
6V
2mA 3.0k 12.0mW
column, the power column, has been attached to R1
7.8V
2mA 3.9k 15.6mW
the tabulated results of the last problem and is R2
shown in Figure 217.
Total Ckt. 13.8V 2mA 6.9k 27.8mW
If you multiply the voltage by the current on a
FIGURE 217
specific row, the product will be the power. Add
the power being dissipated by each resistor and the sum will equal the total power being supplied to the
circuit by the source. That power, for this circuit, is 27.8 milliWatts. There are other formulas that are
available to calculate the power. For the present, however, the formula introduced in Figure 216 will
be sufficient for all power calculations you will be required to perform.
Here is another series circuit problem that requires the use of Ohms Law and the Power formula. You
must remember to use the rules that are applied to the vertical columns, before you attempt a
calculation. In addition, after each calculation you should apply the rules
again.
The circuit is shown in Figure 218.
The problem, as you can see, is a single source series circuit with three
resistors. The values that are GIVEN are as follows: The voltage drop across
resistor R1 is 6.0 Volts and the power being dissipated by R1 is 18
milliWatts. The total circuit power is 60.0 milliWatts and resistor R2 has a
Figure 218
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ELEC 103 UNIT 2
value of 1.5k. The values that you are to
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P
calculate are as follows: The voltage drop across
R1
6.0V
18.0mW
R3, (ER3), and the total circuit resistance (RT).
1.5k
The table for this example is shown in Figure R2
219 with the GIVEN values in their proper R3
*
places, and the values you are to FIND marked
Total Ckt.
*
60.0mW
with an asterisk.
FIGURE 219
What is the first thing you are going to do to start
solving the problem? That is right! Apply the rules to the voltage column, the current column, the
resistance column and the power column. However, for this problem, the rules just do not work the
first time as they did in the previous problem. So, proceed to the next step, which means you are to
look for a row where two pieces of data are available to you. For this example, the first row is the only
row where two pieces of data are available. The next problem you have is to find the proper formula.
However, of the two available, you have only one that has the voltage (E) and the power (P) in the
same formula. Since both the formula we are using have three values, the third value can be calculated
when the other two values are known. The power formula in Figure 216, should be utilized for this
calculation. Since voltage and power are the known values, the current is the unknown value. The
calculation IR1 = PR1  ER1 will involve dividing 18 milliWatts by 6 Volts to calculate the current. The
calculated current, IR1, is 3 mA. Therefore, that value can be placed in the current box on the R1 row.
Before and after each calculation, you must use the rules in association with the columns. The rule for
current applies nicely at this time, and allows you to fill in all the current boxes for the problem with
the 3mA calculated current. The current law states that the current is the same everywhere, in a series
circuit. Therefore, the current through R2, the current through R3 and the total circuit current, IT, are the
same as the current flowing through R1, which was just calculated. Let us review what has occurred so
far. The problem was setup using the tabulation method and the rules were applied to the columns with
no success. The first row had a calculation possibility using the power formula. The current flowing
through R1 was calculated as 3mA. Again, the rules were applied to the table after the calculation was
completed. The rule for current was able to be applied to the current column in the table. If one current
in a series circuit is known, all currents in the series circuit are known. Therefore, 3mA could be placed
in all the current boxes of Figure 220. A
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P
calculation was performed and the calculation
6.0V
3mA
18.0mW
was followed by the application of the rules to R1
3mA 1.5k
the appropriate column. The next step will be to R2
look for a calculation possibility. That means you R3
*
3mA
should look for a row where two pieces of data Total Ckt.
3mA
*
60.0mW
are available. Obviously, there are three rows
FIGURE 220
where that condition exists. The first row, the
second row and the fourth row each have two known values. Which one should be calculated first? It
really makes no difference, since eventually, you will probably fill all the boxes. The second row has
known values for the current (I) and the resistance (R). Therefore, you should pick a formula that has
both the current and the resistance in the formula. Of course, ER2 = IR2 x RR2 is the formula that should
be used for this calculation. The voltage across the resistor R2, (ER2), will be the calculated value. The
voltage drop ER2, is calculated by multiplying 3mA, (IR2), times 1.5k, (R2), for a value of 4.5 Volts.
That value can be placed in the box representing the voltage drop across resistor R2. The rules should
be applied after a calculation. However, nothing can be learned from applying the rules, at this time, so
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ELEC 103 UNIT 2
the procedure continues with another calculation. A calculation could be performed using data on the
first row, the second row or the circuit totals row. Since the fourth row, the circuit totals row, has an
asterisk, it will be used for the calculation. The known values are total circuit power (PT) and total
circuit current (IT). The power formula in Figure 216 will show that voltage equals power divided by
current. The equation should be written with subscripts and should look like this ET = PT  IT.
Substituting into the equation will show 60mW divided by 3mA for a total voltage of 20 Volts.
The calculated voltage should be placed in the
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P
appropriate box and Figure 221 will show all
6.0V
3mA
18.0mW
the circuit information known, up to and R1
4.5V
3mA 1.5k
including the last calculation. If you look at the R2
voltage column of the figure, you will notice all R3
*
3mA
boxes are filled except one. The voltage rule Total Ckt. 20.0V 3mA
*
60.0mW
states, the total voltage equals the sum of the
FIGURE 221
individual voltage drops. Since ET = ER1 + ER2 +
ER3, then transposing the equation will make ER3 = ET  ER1  ER2. Substituting into the equation,
produces a value for ER3 equal to 9.5 Volts. The calculated voltage should be placed in the table in the
proper box and the rules should be applied to the columns. Again, there will be no success when the
rules are applied, so you should look for another calculation to perform. There is one asterisk
remaining in the table and it is in the circuit totals row. There is sufficient information to calculate the
value necessary to fill that box. If that box is filled, the problem is over even though not all the boxes
will be filled. What formula would you use to calculate the total circuit resistance, (RT)?
The only formula with resistance in it is the Ohms Law formula and it should take the form of R T = ET
 IT. Substituting into the equation should show 20V divided by 3mA for a total circuit resistance, R T,
equal to 6.7k. That finishes the calculations necessary to answer the FIND questions set forward in
the original problem statement.
If you want to calculate the information
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necessary to fill the boxes in the table
6.0V
3mA 2.0k
completely, you should do it first before looking R1
4.5V
3mA 1.5k
at Figure 222 for the answers. This is a good R2
time to review the rules by checking the column's R3
9.5V
3mA 3.2k
in the figure to the left. Also, perform some Total Ckt. 20.0V 3mA 6.7k
Ohms Law calculations on various rows and
FIGURE 222
make certain that the numbers in the boxes turn
out correctly when they are multiplied or divided as the case may demand.
P
18.0mW
13.5mW
28.5mW
60.0mW
OHMS LAW AND PARALLEL CIRCUITS
When you studied series circuits, you learned there were three rules associated with series circuits that
you needed to know. If you did not know the rules, in addition to Ohms Law, you could not solve
series circuit problems. The same is true with parallel circuits as was true with series circuits. There are
three rules that apply to parallel circuits and it is necessary for you to learn the three rules before
attempting to solve parallel circuit problems. The rules, as before, are related with voltage, current, and
resistance and can be written in word format and in mathematical format.
1. The rule for voltage states the voltage is the same across all branches of the parallel circuit
and is equal to the applied or source voltage. The applied voltage (VA), the source voltage
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ELEC 103 UNIT 2
2.
3.
(VS), and the total voltage (VT) are all used interchangeably. The mathematical expression would
be written as: VT = VB1 = VB2 = ••• = VBN
The rule for current states the total current flowing in the circuit is equal to the sum of the
individual branch currents. Since there is more than one path for current to flow, it is different
in different parts of the circuit. The mathematical expression would be written as: IT = IB1 + IB2 +
••• + IBN
The rule for resistance states the total resistance
1
of the circuit is smaller than the smallest RT =
1
1
1
+
+ ... +
branch resistance. The mathematical expression
RB1
RB2
RBN
for total circuit resistance is:
The total resistance is equal to the reciprocal of the sum of the reciprocals of the branch resistances, no
matter how many branches there are in the circuit. The above equation can also be manipulated, using
two branches, to produce another equation that is referred to as the product over the sum formula. It
can be used to calculate the total resistance when the circuit consists of only two branches and a source.
Here is the formula:
RB1 x RB2
RT =
RB1 + RB2
Before starting a parallel circuit problem, you should spend some time going over the three rules for
parallel circuits. To help you apply the parallel circuit rules to an actual circuit, refer to Figure 223
which is a parallel circuit with a source and two branches. Each branch has a single resistor, although
there could be more than one resistor in a branch. In this circuit, branch one will be defined as the
vertical space between the points marked "A" and "B" which includes the 6k resistor. Branch two of
the circuit will be defined as the space between the points marked "C" and "D" which includes the 3k
resistor and the milliammeter labeled, Ir. The voltage rule states the voltage across each branch of a
parallel circuit is the same and is equal to the source voltage. Therefore, if a voltmeter was placed
across branch one, with the negative lead placed at point "A" and the positive lead placed at point "B",
the meter would indicate twelve volts. If a voltmeter was placed across branch two, with the negative
lead placed at point "C" and the positive lead placed at point "D", the meter would indicate twelve
volts. In addition, if the voltmeter was placed across the source it would also measure twelve volts. Of
course, you would know where to place the voltmeter leads for that measurement!
The second rule states the total current flowing in the
circuit is equal to the sum of the individual branch
currents. If you read the milliammeter labeled It, in Figure
223, it would read the total circuit current. The reading
would be greater than the milliammeter labeled Ir, since it
is measuring the current in branch two. If there was
another milliammeter placed in series with the 6k
resistor, and the current it measured was added to the
current measured by Ir, it would be equal to the current
measured by the milliammeter labeled, It. The current
Figure 223
flowing through the 6k resistor is the branch one current.
The current flowing through the 3k resistor is the branch two current. The total circuit current can be
measured where the milliammeter, It, is now located or it could be measured between the positive end
of the source and point "B".
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ELEC 103 UNIT 2
To help you put together what has been discussed thus
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far, you should refer to the table shown in Figure 224.
This tabulation of the parallel circuit information will B1
12V
6k
work the same as the tables used with series circuits.
B2
12V
3k
Since the source voltage was given, the rule for voltage
12V
could be applied to the voltage column. That enables Total Ckt
you to fill in all voltage boxes with the same value as
FIGURE 224
the source voltage. In addition, the resistor values can be
obtained from the schematic and placed in the appropriate boxes. If you study the table, you should see
three calculation possibilities. The current flowing in branch one, the current flowing in branch two,
and the total circuit resistance. The total circuit resistance is equal to the reciprocal of the sum of the
reciprocals. However, since there are only two branches, the product over the sum formula can be used.
We will use both formulas to calculate the total circuit resistance. The first formula will be the product
over the sum:
RB1 x RB2
6x103  x 3x103 
18x106  2
RT 
=
=
= 2x103  = 2k
3
3
3
RB1 + RB2
6x10  + 3x10 
9x10 
The next formula is the reciprocal of the sum of the reciprocals formula. You will need a scientific
calculator for this calculation and if you do not have one available, you should locate one before
attempting the calculation. On the calculator keyboard, locate the reciprocal key, which is the key
marked 1/X. The input sequence will be as follows:
6
1/X
+
3
1/X
=
1/X
The number or symbol inside the box is the key that should be pressed on the calculator keyboard.
Since the resistors are in the same k units, the k units were factored out and will be placed after the
answer. If you enter the sequence, as shown, the answer will appear on the calculator display after the
last key has been pressed. Since 2k was calculated earlier as the total circuit resistance, it should be
no different using the reciprocal of the sum of the reciprocals formula. The formula is:
RT 
1
1
1
+
6k
3k
Therefore, you are starting the entry routine in the lower left of the formula and are proceeding just as
the formula directs you to proceed. That is, take the reciprocal of 6k and add that to the reciprocal of 3k
and since that is all the branches being used, hit equals to get a sum of the reciprocals. Take the
reciprocal of the sum to get the total resistance. If you have more than two branches, just extend the
entry routine to include the resistance of the other branches. REMEMBER, if you put all the resistors
into the same units, you can enter just the number into the calculator and attach the units that were
factored out from the resistances, to the answer.
The calculated value for total circuit resistance can be placed in its box. It should now be obvious that
the total circuit resistance is smaller than the smallest branch resistance. There are three current
calculations that can be performed. You will only need to perform two of the calculations though, since
the current rule for parallel circuits will allow you to calculate the third current.
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ELEC 103 UNIT 2
The calculations for current should all be made using Ohms
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Law, in the classic configuration, which is I = E  R. Of
B1
12V
6k
course, you would use the appropriate subscripts with the
12V
3k
formula to insure the correct number values are substituted into B2
the formula. To calculate the current in branch one, the formula Total Ckt 12V
2k
would read IB1 = EB1  RB1, which would mean dividing 12V
FIGURE 225
by 6k, for a branch one current of 2mA. The branch two
current could be calculated in the same manner. Divide 12V, the voltage across branch two, by 3k,
the resistance of branch two, for a branch two current of 4mA.
The current rule for parallel circuits states the total circuit current equals the sum of the branch
currents. In this circuit, the sum of the branch currents is 6mA. Therefore, the total circuit current will
be equal to 6mA. This can be proven quite easily by dividing the source voltage, 12V, by the total
circuit resistance, 2k. This calculation proves the total circuit current is 6mA and is equal to the sum
of the branch currents. As in series circuits, the power can be calculated, using the power formula, and
the total power is equal to the sum of the power being dissipated by the individual circuit components.
The power calculation results and the other circuit
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calculation results are shown in Figure 226. Use the
12V
2mA
24mW
6k
columns of the tabulated results to check the rules for B1
parallel circuits, and do a few calculations across the B2
12V
4mA
48mW
3k
rows using Ohms Law and the power formula.
Total Ckt 12V
6mA
72mW
2k
The voltage is the same across each branch of a
FIGURE 226
parallel circuit and is equal to the voltage applied to
the circuit. Since all the voltages in the voltage column are the same, the voltage law seems to work.
The total circuit current is equal to the sum of the individual branch currents. The two branch currents
do indeed add up to be equal to the total circuit current, so the current rule checks. The total resistance
will be smaller than the smallest branch resistance. The total circuit resistance has a value of 2k and
was calculated two ways for confirmation. The total power being dissipated by the circuit is equal to
the sum of the power being dissipated by each branch. Those are the rules, and if you check the
columns you will see the values in the column boxes obey the parallel circuit rules.
The most difficult rule to understand is the resistance rule. However, you should remember that the
current for each branch is being supplied by the source. If the source must supply current for many
branches, the resistance it sees must be smaller than any branch resistance since as more current flows
the resistance seen by the source must be getting smaller. So each time a branch is added to the parallel
circuit and more current must be supplied by the source, the source sees the total circuit resistance
getting smaller.
Another parallel circuit is shown in Figure 227. Here is a list of the
data that will be utilized to solve the problem.
GIVEN:
FIND:
IB1 = 5mA
RT =
PR1 = 50mW
PR4 =
RB2 = 25k
IB2 =
Figure 227
R2 = 3k
IT =
R3 = 10k
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ELEC 103 UNIT 2
A tabular format has been used for each circuit that has been solved so far, and that will continue in
this problem. However, instead of one table you might want to use additional tables to help you keep
aware of what you know and what you can solve with the rules and the formula. A multiple table
format will be introduced with this problem to demonstrate how useful it can be. There will be the
main circuit table as has been done in the past. However, this time a table will be used for branch one
and a table will be used for branch two. When you look at the circuit in Figure 227, put your thumb
over the resistors R3 and R4. You should be looking at a two resistor series circuit.
Therefore, a table will be created for the series
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Main
circuit that is really branch one. In addition, a
5mA
table will also be created for the series circuit B1
made up of the source and resistors R3 and R4. B2
*
25k
The table for the main circuit is the one on top,
*
*
Figure 228. The table for the series circuit made Total Main
up of the source, R1 and R2, is the middle table,
FIGURE 228
Figure 229. The table for the series circuit made
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up of the source, R3 and R4, is the lower table, B1
Figure 230. The GIVEN values have been R1
50mW
placed in their boxes and the FIND boxes have
R2
3k

been marked.
5mA
The entire TOTAL B1 row of the middle table is Total B1
exactly the same as the B1 row in the main circuit
FIGURE 229
table. Moreover, the TOTAL B2 row of the lower
table is exactly the same as the B2 row in the B2
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main circuit table. An example would be the total
20k
branch one current in Figure 229, of 5mA, can R3
be used as the B1 row current in the main circuit R4


table of Figure 228. In addition, the total branch Total B2
*
25k
two resistances in Figure 230, of 25k, can be
FIGURE 230
used as the B2 row resistance in the main circuit
table of Figure 228.
REMEMBER, use the series circuit rules with the lower two tables and use the parallel circuit
rules with the uppermost table.
The first step is to apply the proper rules to each of the three tables to discover any new information.
The main circuit table provides no new information. However, the middle table, the branch one table,
gives up some information. Since the total branch current is 5mA, the current through R1 and the
current through R2 is the same current. This means all current boxes in the middle table can be filled
with 5mA. The lower table, the branch two tables, also gives up some information. The total resistance
is equal to the sum of the individual resistances so, therefore, the resistance of R4 can be calculated.
Calculate the value of R4.
The second step is to look for calculation possibilities. Two possibilities exist and both are in the
branch one table. You will see them when you insert the information, learned by applying the rules, in
the boxes of the lower two tables. The first calculation will use the power formula and will take the
form; ER1 = PR1  IR1 which means you will be dividing 50mW by 5mA. Write the ER1 voltage in the
appropriate box. The second calculation will use the Ohms Law formula and will take the form; ER2 =
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ELEC 103 UNIT 2
IR2 x RR2 which means you will be multiplying 5mA times 3k. Write the ER2 voltage in the
appropriate box.
The sequence of events that has taken place, while solving this problem, will be reviewed at this time.
You should look at the tables on this page while the review is being presented.
The tables were drawn and the GIVEN values were placed in their appropriate boxes, while the FIND
boxes were marked with an asterisk.
The proper rules were applied to tables, parallel
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Main
circuit rules to the upper table and series circuit
5mA
25V
rules to the middle and lower tables. The B1
information gained from the application of the B2
*
25V
25k

rules is indicated by bold type. The two currents,
*
25V
in the middle table, could be entered into their Total Main
boxes since the current through R1 and R2 is the
FIGURE 231
same as the branch one current, which was given.
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The resistance of R4 could be calculated using the B1
50mW
resistance rule for series circuits. Two R1
10V
5mA
calculations were performed and they produced R2
15V
5mA
3k

the two values that are not bold print. The rules
5mA
25V
were applied to all tables and the middle table Total B1
gave up some information. The total voltage is
FIGURE 232
equal to the sum of the individual voltage drops.
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Therefore, the value that is bold print was placed B2
in its box. Of course, if the voltage across a R3
20k
branch is known, the voltage across all branches
5k

is known. The applied voltage is also known R4
since they are all equal. Therefore, the value of Total B2
*
25V
25k
25V could be placed in all the main circuit
FIGURE 233
voltage boxes and in the box representing the
total voltage across branch two. The main circuit table has a calculation possibility that, if performed,
will fill one of the important FIND boxes. There are other calculation possibilities, but this one in the
main circuit table is one of the most important since the answer brings us one step closer to a solution.
The calculation will use the Ohms Law formula and will take the form; I = E  R which means you will
be dividing 25V by 25mA. Write the branch two current, I , in the appropriate box. Of course, it can be
placed in two boxes. The main circuit table, branch two row current box and the branch two circuit
table, Total B row, current box. Remember, the Total B row in the bottom table and the B2 row in the
main circuit table are exactly the same. After the calculation, apply the rules to the table columns.
The parallel circuit current rule will help us this time. The total circuit current is equal to the sum of the
individual branch currents. This rule will allow you to add the 5mA branch one current to the 1mA
branch two current and obtain the total circuit current. Fill in the appropriate boxes. Another
calculation can be performed using the main circuit table data.
The total resistance can be calculated using Ohms Law and will take the form; RT = E  I which means
you will be dividing 25V by 6mA for a total circuit resistance equal to 4.17k.
There is only one item that remains to be calculated, the power being dissipated by resistor R , P . There
is more than one way to go about calculating that value, so the following procedure is just one way.
Page 17
ELEC 103 UNIT 2
Calculating P involves two calculations following
this method. The first calculation will be to calculate
the voltage across R using the current flowing
through and the resistance of R. E = I x R is the
formula, which means you will be multiplying 1mA
times 5k for a voltage across R equal to 5 Volts.
The second calculation uses the power formula and
takes the form; PR4 = ER4 x IR4 which means you
will be multiplying 5V by 1mA for a power being
dissipated by R4 equal to 5mW.
That completes the solution of the problem even
though not all the boxes have been filled. It is
recommended that you completely fill the boxes in
Figures 234, 235, and 236 so you can gain some
additional experience. The values for each box
appear in the three figures on the next page, but do
the work before checking the next page for answers.
In the meantime, the total circuit resistance will be
calculated using both formulas. The first formula
used will be the product over the sum using the
branch resistances of: RB1 = 5k and RB2 = 25k
E
I
B1
25V
5mA
B2
Total
Main
25V
25V
Main
R
P
1mA
25k

6mA
4.17k
FIGURE 234
E
I
R
B1
R1
10V
5mA
R2
15V
5mA
Total B1
25V
5mA
50mW
3k
B2
FIGURE 235
E
I
R
R3
1mA
20k
R4
1mA
5k
1mA
25k
Total B2
25V
P

P

FIGURE 236
RB1 x RB2
5x103  x 25x103 
125x106  2
RT 
=
=
= 4.17x103  = 4.17k
3
3
3
RB1 + RB2
5x10  + 25x10 
30x10 
The next formula is the reciprocal of the sum of the reciprocals formula. The input sequence will be as
follows:
5
1/X
+
25
1/X
=
1/X
The number or symbol inside the box is the key, or keys that should be pressed on the calculator
keyboard. Since the resistors are in the same k units, the k units were factored out and will be
placed after the answer.
If you enter the sequence, as shown, the answer will appear on the calculator display after the last key
has been pressed. Just add k after the number that appears on the display and you should have the
same answer that was calculated twice
previously.
E
I
R
P
Main
25V 5mA
125mW
The total circuit resistance was calculated using B1
5k
all three methods just to demonstrate the B2
25V 1mA
25k 25mW
methods. It is not necessary to do so in normal
25V
6mA 4.17k 150mW
Total Main
parallel circuit problem solving. The correct
FIGURE 237
values have been placed in all the boxes in the
three figures to the left.
The solution to the problem should be presented as follows:
Page 18
ELEC 103 UNIT 2
GIVEN:
FIND:
E
I
R
P
B1
IB1 = 5mA
RT = 4.17k
R1
10V
5mA
50mW
2k
PR1 = 50mW
PR4 = 5mW
R2
15V
5mA
3k 75mW
RB2 = 25k
IB2 = 1mA
25V
5mA
Total B1
5k 125mW
R2 = 3k
IT = 6mA
FIGURE 238
R3 = 20k
As stated earlier, the table method, demonstrated
E
I
R
P
in the last few problems, is a way of keeping B2
track of information you already know and also R3
20V
1mA 20k 20mW
the data that is available for possible calculations. R4
5V
1mA
5mW
5k
You might not want to use the table method, but
25V
1mA 25k 25mW
Total B2
it is one of the technique's that can be used to
help you learn how to solve series circuit and
FIGURE 239
parallel circuit problems. It makes it easier to
remember the rules and easier to apply the rules to circuit problems. This completes the section on
Ohms Law and Parallel Circuits.
OHMS LAW AND SERIESPARALLEL CIRCUITS
SeriesParallel circuits are, as you most likely have already guessed, a combination of series circuits
and parallel circuits. When working with seriesparallel circuits you must apply the rules for series
circuits to the appropriate series part of the circuit and the rules for parallel circuits to their part of the
seriesparallel circuit. Therefore, instead of working with only three rules you will be working with six
rules. The Ohms Law formula and the power formula are still used as they have been used in the past.
The figure to the right is one of the most simple seriesparallel circuits that can be drawn. It is being
shown to make it easy for you to identify the series portion of the circuit and the parallel portion of the
circuit. When the total circuit current is flowing through a component that is the series circuit portion
of the seriesparallel circuit. When only part of the total circuit current is flowing through a component
that is the parallel circuit portion of the seriesparallel circuit. Referring again to Figure 240, the
resistor R1 would be the series circuit portion of the overall seriesparallel circuit schematic shown in
the figure. All of the circuit current leaving the negative terminal of the voltage source must flow
through resistor R1 to complete the path back to the positive terminal of the voltage source. Therefore,
R1 is identified as the series portion of the seriesparallel circuit. When looking at resistors R2 and R3,
it should be obvious they are in parallel with each other and will each
share some of the current entering the parallel combination. Neither R2
nor R3 will have the total circuit current flowing through them, but will
have just a portion of the total circuit current flowing through them, as
is the usual case in a parallel circuit. Resistors R2 and R3 can be
identified as the parallel circuit portion of the seriesparallel circuit
schematic illustrated in Figure 240.
Figure 240
It is much easier to work with the circuit schematic shown in Figure
241 than it is to work with the circuit schematic shown in the previous figure. The circuit is exactly
the same in both figures.
Page 19
ELEC 103 UNIT 2
However, redrawing the circuit so the schematic presents all the resistors in a vertical configuration
makes it much easier to identify the series and the parallel parts of the circuit. In addition, redrawing
the circuit will make it much easier to solve seriesparallel circuit problems as the circuits become
more complex.
When working on seriesparallel circuit problems, one must learn how to redraw the circuit so the
circuit can be sufficiently simplified so Ohms Law can be applied to the simplified circuit. What is
meant by simplifying the drawing?
A procedure usually requires you to take the following steps.
1.
Redraw the circuit so all components are oriented in a vertical
configuration, similar to Figure 241.
2.
Look as far away from the voltage source as possible and identify one
of two situations. If there is only one resistor located as far away from
the voltage source as possible, it is probably in parallel with another
resistor. If there is more than one resistor located as far away from the
voltage source as possible, it is probably in series with other resistors.
Figure 241
3.
If there is only one resistor, put it in parallel with the other resistor and
calculate the equivalent resistance of the two resistors. Redraw the circuit, replacing the two
resistors with the equivalent resistance in the redrawn circuit.
4.
If there is more than one resistor, add its value to the value of the other resistors in series with
it. Redraw the circuit, replacing those resistors you added together with a resistor of equivalent
value.
5.
Repeat steps 2, 3, and 4 until there is only one resistor and the voltage source remaining in the
circuit. At that point, you will have the total circuit resistance represented by the single
resistance. The total circuit current can then be calculated using the voltage divided by the
resistance of the lone resistor.
An example of the procedure just stated in the five steps can be
demonstrated using the circuit of Figure 240. The circuit is redrawn in
Figure 241 so all components are oriented in a vertical position. Looking as
far away from the voltage source as possible, using Figure 241, you should
see resistor R3 out there all alone.
If you look carefully, you should see resistor R2 is in parallel with resistor
R3. If you use R2 and R3 in the parallel formula for total resistance, you will
calculate a value equivalent to R2 and R3 in parallel. You should redraw the
Figure 242
circuit using all other resistors in the circuit, but replace resistors R2 and R3
with resistor Rx. The redrawn circuit is in Figure 242. The next step is to
apply the rules again to the circuit just redrawn. The result of applying the
rules to the circuit in Figure 242 is the circuit in Figure 243.
Of course, the schematic shown in Figure 243 is a series circuit and series
circuits can be solved very easily, at this point in your experience, if
component values are made available. Well, let us assign values to the
components in Figure 240 and work through a typical seriesparallel
circuit problem.
Figure 243
Using Figure 240 as the basic seriesparallel circuit schematic, let the
Page 20
ELEC 103 UNIT 2
source be VS = 30 Volts. The resistors in the circuit will be assigned the following values:
GIVEN values: R1 = 1.8k, R2 = 2K, and R3 = 3k.
You should FIND the current through R3, (IR3), and the voltage dropped across R1, (VR1).
A table can be used to help you solve seriesparallel circuit problems, but, with all seriesparallel
circuit problems, the only use the table has is to keep a record of component values and to provide the
proper data for use with Ohms Law. The rules can no longer be applied to the columns as they were
with series circuits and with parallel circuits. The column's are a mixture of values and will be unusable
for the application of the rules. On the next page you will see the GIVEN and FIND values, the table
that will be used to record the information that is given, and calculated. In addition, you will see the
three circuit schematics that will be necessary to solve this seriesparallel circuit problem.
In the meantime, an explanation of how the table is setup and how it will be used might be beneficial.
The table is somewhat similar to the previous tables you have seen and used. This time, in addition to
the rows for the circuit resistances, a row is added for each resistor that is created by you as you reduce
the circuit down to just the source and one resistor. The last circuit with one resistor is, in effect, equal
to the total circuit resistance.
The first step to take in solving the seriesparallel circuit
E
I
R
problem is to look at the original schematic, which is R1
*
1.8k
shown in Figure 240. Next, setup the table as you see to
2.0k
the left. If you do not know how many extra rows to add, R2
*
3.0k
leave room for one or two with simple circuits and more R3
with complex circuits. The second step is to redraw the RX
original circuit so all the resistors in the circuit are
TOTAL
30V
orientated in a vertical configuration. The original circuit
FIGURE 244
has been redrawn in Figure 245. Redrawing the original
circuit does not seem to make sense for such a simple circuit. However, redrawing the circuit will
definitely make circuit problems easier to solve, as the circuits become more complex. After you
redraw the circuit, you should look as far away from the voltage source as possible.
FIGURE 245
FIGURE 246
FIGURE 247
This circuit, when it has been redrawn has only one resistor in the position as far away from the source
as possible. That resistor is obviously, R3. Therefore, if you look for what is in parallel with resistor R3,
you should see resistor R2 is in parallel with R3. The next step is to use the reciprocal of the sum of the
reciprocals formula, or the product over the sum formula to calculate the equivalent resistance of the
two resistors. Using the values R2 = 3k and R3 = 2k, one will calculate an equivalent resistance of
1.2k when using either formula. This equivalent resistance should be given a new name, different
from any other resistor in the circuit. This resistor is labeled RX, but could have been labeled anything
you wish. This resistor, RX, gets a row in the table and will be treated as a real resistor in the circuit of
Page 21
ELEC 103 UNIT 2
Figure 246. The value for RX, that was just calculated, should be entered into the table. The next step
is to look as far away from the voltage source as possible, and check for one of the two possible
situations discussed earlier. The second situation exists, so the values of the resistors, when added, will
be represented by a single resistor. Since this will be the only resistor in the circuit, it will be given the
name RT, thereby representing the total circuit resistance. The following is a summation of the
calculations required for the circuit in Figure 247.
The total circuit current, It, for the circuit in Figure 247
E
I
R
can be calculated using ET = 30V and RT = 3.0k. The
*
1.8k
calculated total circuit current is entered in the table to the R1
left. Since there is only one resistor in the circuit of Figure R2
2.0k
247, nothing else can be calculated concerning the circuit R3
*
3.0k
(Power has not been requested). Since everything about the
1.2k
circuit is known, you should move to the previous RX
30V
10mA
3.0k
schematic. In this problem, move to Figure 246 where TOTAL
you will determine everything you can about the circuit
FIGURE 248
shown there. Since the two circuits, the circuit of Figure
246 and the circuit of Figure 247, are equivalent circuits, what you know about one circuit you know
about both circuits.
Since the current flowing in the circuit of Figure 247 is R1
*
10mA
1.8k
10mA, the current flowing in the circuit of Figure 246
2.0k
must also be 10mA. If that is true, then the current flowing R2
*
3.0k
through R1 and the current flowing through RX must both R3
be 10mA, since the circuit is a series circuit. Put that value RX
10mA
1.2k
in the table to the left for both resistors as is shown. The
TOTAL
30V
10mA
3.0k
usual procedure is to determine everything about a circuit
FIGURE 249
that is possible, before moving on to the previous circuit. If
the current through both the resistors is known and the resistance of the resistors is known, the voltage
drops across the resistors can be calculated. Look at the table in Figure 249. The rows for R1 and RX
have unknown values for voltage and known values for current and resistance. If the voltage drops for
resistor R1, (ER1), and for resistor RX, (ERX), are calculated, you then will know everything about the
circuit shown in Figure 246. You can then move on to the previous circuit, the circuit shown in Figure
245. Enter the voltage drops you calculated for ER1 and for ERx into the appropriate boxes of the table
in Figure 249. Since the resistors R1 and RX are in a series circuit, their voltage drops should add up
to be equal to the applied voltage of 30 Volts. Does that happen? When you move to the circuit in
Figure 245, you should be aware that the resistor RX in Figure 246 was representing the parallel
combination of R2 and R3 in this circuit. If RX was representing the parallel combination of R2 and R3,
then the voltage across RX must be the voltage across the parallel combination of R2 and R3.
Additionally, the sum of the currents flowing through R2 and R3 must be equal to the current flowing
through RX. Enter the voltage drops into the appropriate boxes in the table of Figure 249. After
entering the voltages in the table, it should be obvious the only calculations remaining are the two
current calculations. It is easy to make those calculations and add them together to check if they equal
the current flowing through RX. The boxes should all be filled and the results can be seen in Figure
250.
Page 22
ELEC 103 UNIT 2
The procedure you have followed to solve this problem R1
18V
10mA
1.8k
will be used to solve more complex problems in other
R2
12V
6mA
2.0k
courses. Therefore, it might be beneficial if you spend
12V
4mA
3.0k
some time reviewing the procedure used in solving the R3
problem. Also, review the calculations that were RX
12V
10mA
1.2k
performed along the way, and make certain you
TOTAL
30V
10mA
3.0k
understand how the rules for series circuits and the rules
FIGURE 250
for parallel circuits were applied in this problem.
Homework problems for this unit are listed on the next page. Please submit the homework on 8½ x 11
paper with no ragged edges. Draw all circuit schematics and show all work. YOUR HOMEWORK
WILL NOT BE ACCEPTED UNLESS THESE CONDITIONS ARE MET. HOMEWORK
SHOULD BE SUBMITTED TO YOUR PROFESSOR.
Page 23
ELEC 103 UNIT 2
HOMEWORK PROBLEMS FOR UNIT TWO
1. Draw the circuit and calculate the current that flows through four series connected resistors when
R1 = 150, R2 = 250, R3 = 125, R4 = 75 and the source voltage is 12 Volts.
2. Draw the circuit for three series connected resistors supplied by a 25 Volt source. Calculate the
voltage drop across each resistor and the power dissipated by each resistor. The resistor values
are; R1 = 6k, R2 = 13k, and R3 = 11k.
3. A small transistor radio normally draws 15mA from a 9 Volt battery. Calculate the value of the
resistor that must be connected in series with the radio if it is to be supplied from a 12 Volt
battery. Also, calculate the power rating for the series resistor. HINT: Draw the circuit schematic.
4. Draw the circuit of four parallel connected resistors being supplied by a 25 Volt source. Three of
the resistors have values of R1 = 1k, R2 = 12k, and R3 = 8.2k. If the total circuit current is
measured as 36.5mA, determine the value of the fourth resistor.
5. Three parallel connected resistors; R1 = 1.2k, R2 = 3.9k, and R3 = 4.7k are supplied by a
voltage source with an unknown value. What is the total circuit resistance?
6. Two resistors, connected in parallel, draw a total of 75mA from the voltage source. R1 = 620
and R2 = 880. Calculate IR1, IR2 and the source voltage. Also, determine the power dissipated
by each resistor.
7.
Analyze the circuit shown in Figure 251 to determine all resistor currents and voltages. Also,
determine the total circuit current and the power dissipated by
resistors R1 and R3. VS = 9 Volts, R1 = 3.3k, R2 = 12k, R3
= 6.8k, and R4 = 18k.
8.
If resistor R4 is open circuited in the circuit of Figure 251,
what is the value of the total circuit current that is flowing
under that condition?
9.
If resistor R4 is short circuited in the circuit of Figure 251,
FIGURE 251
what is the value of the total circuit current that is flowing
under that condition?
10. Draw the circuits for problems number 8 and number 9. Label all components in each circuit.
Page 24