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Topic 5: Electric currents
5.2 Electric circuits
5.2.1 Define electromotive force.
5.2.2 Describe the concept of internal
resistance.
Topic 5: Electric currents
5.2 Electric circuits
Define electromotive force.
The electromotive force  (or emf) of a cell is
the amount of chemical energy converted to
electrical energy per unit charge.
Since energy per
unit charge is
volts, emf is
measured in volts.
This battery has
an emf of
 = 1.6 V.
Because the
battery is not
connected to any
circuit we say
that it is unloaded.
01.6
00.0
Topic 5: Electric currents
5.2 Electric circuits
Define electromotive force.
EXAMPLE:
How much chemical energy is converted to
electrical energy by the cell if a charge of
15 C is drawn by
the voltmeter?
01.6
SOLUTION:
From ∆V = ∆EP/q
we have  = ∆EP/q.
Thus
∆EP = q
= (1.6)(1510-6)
= 2.410-5 J.
Topic 5: Electric currents
5.2 Electric circuits
Describe the concept of internal resistance.
If we connect a resistor as part of an external
circuit, we see that the voltage read by the
meter goes down a little bit.
We say that the
battery is loaded
because there is
a resistor connected externally
in a circuit.
This battery has
a loaded
potential
difference (p.d.)
of V = 1.5 V.
01.5
01.6
Topic 5: Electric currents
5.2 Electric circuits
Describe the concept of internal resistance.
All cells and batteries have limits as to how
rapidly the chemical reactions can maintain the
voltage.
There is an internal resistance r associated with
a cell or a battery which causes the cell’s
voltage to drop when there is an external demand
for the cell’s electrical energy.
The best cells will have small internal
resistances.
The internal resistance of a cell is why a
battery becomes hot if there is a high demand for
current from an external circuit.
Cell heat will be produced at the rate given by
P = I2r
r = internal resistance
cell heat rate
Topic 5: Electric currents
5.2 Electric circuits
Describe the concept of internal resistance.
PRACTICE: A battery has an internal resistance of
r = 1.25 . What is its rate of heat production
if it is supplying an external circuit with a
current of I = 2.00 A?
SOLUTION:
Rate of heat production is power.
P = I2r
P = (22)(1.25) = 5.00 J/s (5.00 W.)
FYI
If you double the current, the rate of heat
generation will quadruple because of the I2
dependency.
If you accidentally “short circuit”
a battery, the battery may even heat
up enough to leak or explode!
Topic 5: Electric currents
5.2 Electric circuits
Describe the concept of internal resistance.
If we wish to consider the internal
resistance of a cell, we can use
the cell and the resistor symbols

together, like this:
And we may enclose the whole cell in
a box of some sort to show that it is
one unit.
Suppose we connect our cell with its
internal resistance r to an external
circuit consisting of a single resistor
R
of value R.
All of the chemical energy from the
battery is being consumed by the
internal resistance r and the external
resistance R.
r

r
Topic 5: Electric currents
5.2 Electric circuits
Describe the concept of internal resistance.
But from EP = qV we can write:
Electrical energy being created from
chemical energy in the battery: EP = q.

R
Electrical energy being converted to
heat energy by R: EP,R = qVR.
r
Electrical energy being converted to
heat energy by r: EP,r = qVr.
From conservation of energy EP = EP,R + EP,r so
that q = qVR + qVr.
Note that the current I is the same everywhere.
From Ohm’s law VR = IR and Vr = Ir so that
q = qIR + qIr.
 = IR + Ir = I(R + r)
emf relationship
Topic 5: Electric currents
5.2 Electric circuits
1.6 V
1.5 V
330 
Describe the concept of internal resistance.
EXAMPLE:
A cell has an unloaded voltage of 1.6 V
and a loaded voltage of 1.5 V when a

330  resistor is connected as shown.
(a) Find .
r
(b) Then from the second circuit find I.
(c) Finally, find the cell’s internal resistance.
SOLUTION:
(a) From the first schematic we see

that  = 1.6 V.
(Unloaded cell.) R
(b) From the second diagram we see
r
that the voltage across the 330 
resistor is 1.5 V.
V = IR so that 1.5 = I(330) and I = 0.0045 A.
Topic 5: Electric currents
5.2 Electric circuits
1.6 V
1.5 V
330 
Describe the concept of internal resistance.
EXAMPLE:
A cell has an unloaded voltage of 1.6 V
and a loaded voltage of 1.5 V when a

330  resistor is connected as shown.
(a) Find .
r
(b) Then from the second circuit find I.
(c) Finally, find the cell’s internal resistance.
SOLUTION:
(c) Now we can use the emf

R
relationship  = IR + Ir.
1.6 = (0.0045)(330) + (0.0045)r
r
1.6 = 1.5 + 0.0045r
0.1 = 0.0045r
r = 22 .
Topic 5: Electric currents
5.2 Electric circuits
5.2.3 Apply the equations for resistors in series
and parallel.
5.2.4 Draw circuit diagrams.
5.2.5 Describe the use of ideal ammeters and
ideal voltmeters.
Topic 5: Electric currents
5.2 Electric circuits
Apply the equations for resistors in series and
parallel.
Resistors can be connected to one another in
series, which means one after the other.
R1
R2
R3

Note there is only one current I and I is the
same for all components of a series circuit.
Conservation of energy tells us  = V1 + V2 + V3.
Thus  = IR1 + IR2 + IR3
(from Ohm’s law)
 = I(R1 + R2 + R3)
(factoring out I)
 = I(R), where R = R1 + R2 + R3.
R = R1 + R2 + R3
equivalent resistance in series
Topic 5: Electric currents
5.2 Electric circuits
Apply the equations for resistors in series and
parallel.
EXAMPLE: Three resistors
R3
R2
of 330  each are
R1
connected to a 6.0 V battery in series as shown.

(a) What is the circuit’s equivalent resistance?
(b) What is the current in the circuit?
(c) What is the voltage on each resistor?
SOLUTION:
(a) In series, R = R1 + R2 + R3 so that
R = 330 + 330 + 330 = 990 .
(b) Since the voltage on the entire circuit is
6 V, and since the total resistance is 990 ,
from Ohm’s law we have
I = V/R = 6/990 = 0.0061 A.
Topic 5: Electric currents
5.2 Electric circuits
Apply the equations for resistors in series and
parallel.
EXAMPLE: Three resistors
R3
R2
of 330  each are
R1
connected to a 6.0 V battery in series as shown.
(a) What is the circuit’s equivalent resistance?
(b) What is the current in the circuit?
(c) What is the voltage on each resistor?
SOLUTION:
FYI
(c) The current I we just found is
In general
the same everywhere. Thus each resis- the V’s
tor has a current of I = 0.0061 A.
are
different
From Ohm’s law, each resistor has a
if the R’s
voltage given by
are.
V = IR = (.0061)(330) = 2.0 V.
Topic 5: Electric currents
5.2 Electric circuits
Apply the equations for resistors in series and
parallel.
Resistors can also be

connected in parallel.
R2
R3
R1
Each resistor is connected
directly to the cell.
Thus each resistor has the same voltage V and V
is the same for all components of a parallel
circuit. We can then write  = V1 = V2 = V3  V.
But there are three currents I1, I2, and I3.
Since the total current I passes through the cell
we see that I = I1 + I2 + I3.
If R is the equivalent or total resistance of the
three resistors, then I = I1 + I2 + I3 becomes
/R = V1/R1 + V2/R2 + V3/R3
(Ohm’s law I = V/R)
Topic 5: Electric currents
5.2 Electric circuits
Apply the equations for resistors in series and
parallel.
Resistors can also be

connected in parallel.
R2
R3
R1
Each resistor is connected
directly to the cell.
Thus each resistor has the same voltage V and V
is the same for all components of a parallel
circuit. We can then write  = V1 = V2 = V3  V.
From
 = V1 = V2 = V3  V
and
/R = V1/R1 + V2/R2 + V3/R3
we get
V/R = V/R1 + V/R2 + V/R3.
Thus the equivalent resistance R is given by
1/R = 1/R1 + 1/R2 + 1/R3
equivalent resistance
in parallel
Topic 5: Electric currents
5.2 Electric circuits
Apply the equations for resistors in series and
parallel.
EXAMPLE: Three resistors
of 330  each are

R2
R3
R1
connected to a 6.0 V cell
in parallel as shown.
(a) What is the circuit’s equivalent resistance?
(b) What is the voltage on each resistor?
(c) What is the current in each resistor?
SOLUTION:
(a) In parallel, 1/R = 1/R1 + 1/R2 + 1/R3 so that
1/R = 1/330 + 1/330 + 1/330 = 0.00909.
Thus R = 1/0.00909 = 110 .
(b) The voltage on each resistor is 6 V, since
the resistors are in parallel. (Or each resistor
is clearly directly connected to the battery).
Topic 5: Electric currents
5.2 Electric circuits
Apply the equations for resistors in series and
parallel.
EXAMPLE: Three resistors
of 330  each are

R2
R3
R1
connected to a 6.0 V cell
in parallel as shown.
(a) What is the circuit’s equivalent resistance?
(b) What is the voltage on each resistor?
(c) What is the current in each resistor?
SOLUTION:
FYI
(c) Using Ohm’s law (I = V/R):
I1 = V1/R1 = 6/330 = 0.018 A. In general the
I’s are
I2 = V2/R2 = 6/330 = 0.018 A.
different if
I3 = V3/R3 = 6/330 = 0.018 A. the R’s are
different.
Topic 5: Electric currents
5.2 Electric circuits
Draw circuit diagrams.
A complete circuit will always contain a cell or
a battery.
The schematic diagram of a cell is this:
A battery is just a group
of cells connected in series:
If each cell is 1.5 V, then
the battery shown above is
3(1.5) = 4.5 V.
Often a battery is shown in
a more compact form:
Of course a fixed-value
resistor looks like this:
cell
battery
battery
resistor
Topic 5: Electric currents
5.2 Electric circuits
Draw circuit diagrams.
EXAMPLE: Draw schematic diagrams of each of the
solder joints
following circuits:
A
SOLUTION:
For A:
B
For B:
Topic 5: Electric currents
5.2 Electric circuits
Describe the use of ideal ammeters and ideal
voltmeters.
PRACTICE: Draw a schematic diagram
for this circuit:
SOLUTION:
FYI
Be sure to position the
voltmeter across the correct
resistor in parallel.
1.06
Topic 5: Electric currents
5.2 Electric circuits
Describe the use of ideal ammeters and ideal
voltmeters.
PRACTICE: Draw a schematic diagram
for this circuit:
SOLUTION:
.003
FYI
Be sure to position the
ammeter between the correct
resistors in series.
Topic 5: Electric currents
5.2 Electric circuits
Draw circuit diagrams.
PRACTICE: Draw a schematic diagram
for this circuit:
SOLUTION:
FYI
This circuit is a combination series-parallel.
In a later slide you will learn how to find the
equivalent resistance of the combo circuit.
Topic 5: Electric currents
5.2 Electric circuits
Describe the use of ideal ammeters and ideal
voltmeters.
We have seen that voltmeters
are always connected in
parallel.
The voltmeter reads the voltage
of only the component it is in
parallel with.
The green current represents the amount of
current the battery needs to supply to the
voltmeter in order to make it register.
The red current is the amount of current the
battery supplies to the original circuit.
In order to NOT ALTER the original properties of
the circuit, ideal voltmeters have extremely high
resistance () to minimize the green current.
Topic 5: Electric currents
5.2 Electric circuits
Describe the use of ideal ammeters and ideal
voltmeters.
We have seen that ammeters
are always connected in
series.
The ammeter reads the
current of the original circuit.
In order to NOT ALTER the original properties of
the circuit, ideal ammeters have extremely low
resistance (0) to minimize the effect on the red
current.
Topic 5: Electric currents
5.2 Electric circuits
5.2.6 Describe a potential divider.
5.2.7 Explain the use of sensors in potential
divider circuits.
5.2.8 Solve problems involving electric circuits.
Topic 5: Electric currents
5.2 Electric circuits
IN
Describe a potential divider.
Consider a battery of  = 6 V. Suppose we have a
light bulb that can only use three volts. How do
we obtain 3 V from a 6 V battery?
A potential divider is a circuit made of two (or
more) series resistors that allows us to tap off
any voltage we want that is
potential
less than the battery voltage.
divider
R
The input voltage is the
1
emf of the battery.
The output voltage is the
voltage drop across R2.
R2
Since the resistors are in
series R = R1 + R2.
OUT
Topic 5: Electric currents
5.2 Electric circuits
IN
Describe a potential divider.
Consider a battery of  = 6 V. Suppose we have a
light bulb that can only use three volts. How do
we obtain 3 V from a 6 V battery?
A potential divider is a circuit made of two (or
more) series resistors that allows us to tap off
any voltage we want that is
potential
less than the battery voltage.
divider
R
1
From Ohm’s law the current I
of the divider is given by
I = VIN/R = VIN/(R1 + R2).
But VOUT = V2 = IR2 so that
R2
VOUT = IR2
= R2VIN/(R1 + R2).
OUT
VOUT = VIN[ R2/(R1 + R2) ]
potential divider
Topic 5: Electric currents
5.2 Electric circuits
IN
Solve problems involving electric circuits.
PRACTICE:
Find the output voltage if
R1
the battery has an emf of
9.0 V, R1 is a 2200 
resistor, and R2 is a 330 
resistor.
R2
SOLUTION:
Use VOUT = VIN[ R2/(R1 + R2) ]
VOUT = 9[ 330/(2200 + 330) ]
VOUT = 9[ 330/2530 ]
VOUT = 1.2 V.
FYI
The bigger R2 is in comparison to R1, the bigger
VOUT is in proportion to the total voltage.
OUT
Topic 5: Electric currents
5.2 Electric circuits
IN
Solve problems involving electric circuits.
PRACTICE:
Find the value of R2 if
R1
the battery has an emf of
9.0 V, R1 is a 2200 
resistor, and we want an
R2
output voltage of 6 V.
SOLUTION:
Use the formula VOUT = VIN[ R2/(R1 + R2) ]. Thus
6 = 9[ R2/(2200 + R2) ]
6(2200 + R2) = 9R2
13200 + 6R2 = 9R2
13200 = 3R2
R2 = 4400 
OUT
Topic 5: Electric currents
5.2 Electric circuits
Explain the use of sensors in potential divider
circuits.
PRACTICE: A light-dependent
R1
resistor (LDR) has a resistance
9.0 V
of 25  in bright light and a
resistance of 22000  in low
light. An electronic switch
electronic
R
will turn on a light when its
2
switch
p.d. is above 7.0 V. What
should the value of R1 be?
SOLUTION: Use VOUT = VIN[ R2/(R1 + R2) ]. Thus
7 = 9[ 22000/(R1 + 22000) ]
7(R1 + 22000) = 9(22000)
7R1 + 154000 = 198000
7R1 = 44000
R1 = 6300 
(6286)
Topic 5: Electric currents
5.2 Electric circuits
Explain the use of sensors in potential divider
circuits.
PRACTICE: A thermistor has a
R1
resistance of 250  when it
9.0 V
is in the heat of a fire and
a resistance of 65000  when
at room temperature. An
electronic
R
electronic switch will turn
2
switch
on a sprinkler system when
its p.d. is above 7.0 V.
(a) Should the thermistor be R1 or R2?
SOLUTION:
Because we want a high voltage at a high
temperature, and because the thermistor’s
resistance decreases with temperature, it should
be placed at the R1 position.
Topic 5: Electric currents
5.2 Electric circuits
Explain the use of sensors in potential divider
circuits.
PRACTICE: A thermistor has a
R1
resistance of 250  when it
V
9.0 V
is in the heat of a fire and
1
a resistance of 65000  when
at room temperature. An
electronic
R
electronic switch will turn
2
switch
on a sprinkler system when
its p.d. is above 7.0 V.
(b) What should the value of R2 be?
V
SOLUTION: In fire the thermistor is R1 = 250 . 2
7 = 9[ R2/(250 + R2) ]
7(250 + R2) = 9R2
1750 + 7R2 = 9R2
R2 = 880  (875)
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: A filament lamp is
X
rated at “4.0 V, 0.80 W” on its
Y
package. The potentiometer has a
resistance from X to Z of 24 
Z
and has linear variation.
7.0 V
(a) Sketch the variation of the
p.d. V vs. the current I for a
typical filament lamp. Is it ohmic?
ohmic means linear
SOLUTION:
Since the temperature increases with
V
the current, so does the resistance.
non-ohmic
But from V = IR we see that R = V/I,
which is the slope.
Thus the slope should increase with I.
I
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: A filament lamp is
X
rated at “4.0 V, 0.80 W” on its
Y
package. The potentiometer has a
resistance from X to Z of 24 
Z
and has linear variation.
7.0 V
(b) The potentiometer is adjusted
so that the meter shows 4.0 V.
Will it’s contact be above Y,
below Y, or exactly on Y?
SOLUTION:
The circuit is acting like a potential divider
with R1 being the resistance between X and Y and
R2 being the resistance between Y and Z.
Since we need VOUT = 4 V, and since VIN = 6 V, the
contact must be adjusted above the Y.
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: A filament lamp is
X
rated at “4.0 V, 0.80 W” on its
Y
package. The potentiometer has a
resistance from X to Z of 24 
Z
and has linear variation.
7.0 V
(c) The potentiometer is adjusted
so that the meter shows 4.0 V.
What are the current and the
resistance of the lamp at this
instant?
SOLUTION:
P = 0.80 and V = 4.0.
From P = IV we get 0.8 = I(4) so that I = 0.20 A.
From V = IR we get 4 = 0.2R so that R = 20. .
You could also use P = I2R for this last one.
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: A filament lamp is
X
rated at “4.0 V, 0.80 W” on its
R1
Y
package. The potentiometer has a
R2
resistance from X to Z of 24 
Z
and has linear variation.
7.0 V
(d) The potentiometer is adjusted
so that the meter shows 4.0 V.
What is the resistance of the Y-Z
portion of the potentiometer?
SOLUTION:
Let R1 = X-Y resistance, R2 = Y-Z resistance.
Then R1 + R2 = 24 so that R1 = 24 – R2.
From VOUT = VIN[ R2/(R1 + R2) ] we get
4 = 7[ R2/(24 – R2 + R2) ].
R2 = 14 .
(13.71)
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: A filament lamp is
X
rated at “4.0 V, 0.80 W” on its
R1
Y
package. The potentiometer has a
R2
resistance from X to Z of 24 
Z
and has linear variation.
7.0 V
(e) The potentiometer is adjusted
so that the meter shows 4.0 V.
What is the current in the Y-Z
portion of the potentiometer?
SOLUTION:
V2 = 4.0 V because it is in parallel with the
lamp.
Then I2 = V2/R2
= 4/13.71 (use unrounded value)
= 0.29 A
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: A filament lamp is
X
rated at “4.0 V, 0.80 W” on its
R1
Y
package. The potentiometer has a
R2
resistance from X to Z of 24 
Z
and has linear variation.
7.0 V
(f) The potentiometer is adjusted
so that the meter shows 4.0 V.
What is the current in the
ammeter?
SOLUTION:
There are two currents supplied by the battery.
The red current is 0.29 A because it is the I2 we
just calculated in (e).
The green current is 0.20 A found in (c).
The ammeter has both so I = 0.29 + 0.20 = 0.49 A.
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: Which circuit shows the correct setup
to find the V-I characteristics of a filament
lamp?
lamp
two
SOLUTION:
current
currents
The voltmeter
must be in parallel with the lamp.
It IS, in ALL
cases.
no
short
currents
circuit!
The ammeter must
be in series with
the lamp and must
read only the
lamp’s current.
The correct response is B.
Topic 5: Electric currents
5.2 Electric circuits
equivalent ckt
Solve problems involving electric circuits.
PRACTICE: A non-ideal voltmeter is used to
measure the p.d. of the 20 k resistor as
shown. What will its reading be?
SOLUTION:
There are two currents passing
through the circuit because the
voltmeter does not have a high
enough resistance to prevent the
green one from flowing.
The 20 k resistor is in parallel with the 20 k
voltmeter. Thus their equivalent resistance is
1/R = 1/20000 + 1/20000 = 2/20000.
R = 20000/2 = 10 k.
But then we have two 10 k resistors in series
and each takes half the battery voltage, or 3 V.
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: All three circuits use the same
resistors and the same cells.
Highest I
0.5R
Lowest I
2R
parallel
series
Which one of the following
shows the correct ranking
for the currents passing
through the cells?
SOLUTION:
The bigger the resistance
the lower the current.
Middle I
1.5R
R
0.5R
combo
Topic 5: Electric currents
5.2 Electric circuits
Solve problems involving electric circuits.
PRACTICE: The battery has
E
negligible internal resistance and the voltmeter
has infinite resistance.
What are the readings on
the voltmeter when the
switch is open and closed?
SOLUTION:
With the switch open the
green R is not part of the
circuit. Red and orange
split the battery emf.
With the switch closed the
red and green are in parallel and are (1/2)R.