Download Some Useful Electrical Circuit Information

Some Useful Electrical Circuit Information
An Analogy .. the flow of water (e.g. in a central-heating system)
Voltage corresponds to Pressure.
Electric Charge corresponds to the volume of water. When it starts to flow, it
becomes ..
Electric Current which corresponds to the Volume Flow Rate of the water.
Circuit Principles
A current will only flow if it has a complete circuit.
Ohm's Law: V = I  R, where V = voltage, I = current, and R = resistance. The
units are volt (V), ampere (A) and ohm (often spelled out but also often abbreviated
to the Greek ) respectively. Note that it follows by Sums that I = V/R and R = V/I.
Kirchhoff's Laws:
1.
The total current flowing into any junction in a circuit is equal to the total
current flowing out of the junction.
2.
Round any complete loop in a circuit, the sum of the voltages generated by
voltage generators (e.g. batteries or the mains) is equal to the sum of the products of
current times resistance for all the resistances in the loop.
This circuit should illustrate the idea.
I1
3 ohm
+
3V
-
4 ohm
I2
+
5 ohm
4V
-
I3
The currents I1 and I2 are flowing into the junction at the top, whilst I3 is flowing out of
it .. so I1 + I2 = I3.
Going round the left-hand loop clockwise, we encounter one voltage (3 V) and two
resistances of 3 and 5 ohm respectively .. so 3 = 3I1 + 5I3. Likewise anticlockwise
round the right-hand loop .. 4 = 4I2 + 5I3. We can solve the three equations
"simultaneously" to obtain I1, I2 and I3 (they tell me it is easy to do so if we use a TI85 or something). We may revisit such situations later on when we are more in
practice but we will keep things simpler for the moment. But we can apply Kirchhoff
with a minimum of sums to see what will happen in the following cases:
3 ohm
I
(a
)
4 ohm
+
-
5
ohm
3V
-
ohm
5
3V
+
(b)
-
+
1.5 V
I
In Circuit (a), we just have one loop, so we first add up the voltages; an easy task as
there is just one of them … 3 V. Then we look at the deadly “IR Drops” and see to
our horror that there are three of them, but at least the current I is the same in each
(why ?). So we have to add up 3 x I for the 3-ohm, 4 x I for the 4-ohm and 5 x I for
the 5-ohm, giving (3 + 4 + 5) x I or 12 I. Mr. (Herr Doktor ?) Kirchhoff says that these
two answers are equal, so 12 x I = 3 and I = 3/12 = 0.25 A.
We could have got the same answer by just adding up the resistances and using
Ohm’s Law (see below under Resistances in Series)
In Circuit (b), we have two voltage sources and only one resistance. In this case, we
add up the voltages, noting that both are driving the current the same way round the
circuit; 1.5 + 3 = 4.5 V. There is only one resistance, so the “sum of IR drops” is just
the IR drop, or 5 x I. So 5 x I = 4.5 and I = 4.5/5 = 0.9 A.
So it looks as if we can add up the voltages of generators in series in such a
situation.
Do we know any practical situations in which we might have two voltage sources
(e.g. batteries) in series, either the same way round or opposing each other ?
Resistances in Series and Parallel
Series:
R1
R2
R3
Total Resistance = R1 + R2 + R3
Parallel :
R1
R2
Total R is calculated from
R3
1
1
1
1



R R1 R2 R3 .
Power in Electrical Circuits
Power = Voltage x Current (applies generally)
Power = V2/R = I2R (only applies to resistances)
Internal Resistance
We find that all practical sources of voltage behave as if they have some resistance
in series with their terminals. The practical effect of this (at least partly imaginary)
resistance is that the terminal voltage of the source falls off as the current it is
supplying increases. A frustrating manifestation is that of a failing car battery which
may well be giving 12 V when no current is drawn but, when we operate the starter,
we hear an ominous "rr .. rr .. rr .." instead of "rrrrrrrVROOM". We can represent
such a practical voltage source by an ideal one in series with the resistance (an
Equivalent Circuit).
Electromotive force (e.m.f.) E
Internal resistance R
First Electrical Examples Sheet
1.
State and explain Ohm's Law and Kirchhoff's laws.
2.
Complete the following table.
V, volt
10
240
?
15
110
?
I, ampere
2.5
?
3
3
?
17
R, ohm
?
48
20
?
5.5
60
What is the power in each of the above cases ? (25, 1200, 180, 45, 2200, 17340 w)
3.
What is the total resistance of three 30-ohm light-bulbs if they are connected
(a) in series, (b) in parallel ?
(90 ohm, 10 ohm)
4. If the above lamps are intended to work at 24 volt, what will be their power rating?
If we connected some of them in series to a 240-volt power supply, how many would
we have to connect in series if their correct current was to flow ? (19.2 watt, 10
lamps)
5. If we connect two 60-watt lamps and a 1-kw electric fire in parallel across the 240volt mains, what total current will be drawn ?
(4.67 A)
6.
2
ohm
+
12
ohm
20 volt
-
24
ohm
Determine the current in each resistor. (Hint: work out the parallel resistance of the
12- and 24-ohm combination and use it to work out the total series resistance and
therefore the current in the 2-ohm resistance. Then continue to use Ohm's Law
again to calculate the voltage across the parallel combination .. and therefore the
currents)
(2 A, 1.333 A, 0.667A)
7. This circuit represents a system in which a light at a remote road junction can be
supplied either by a 24-volt battery of internal resistance 4 ohm, or by a wind
generator of internal resistance 2 ohm which generates a voltage of 2u volt on opencircuit where u is the wind speed in km/hr. The generator also serves to keep the
battery charged.
(a)
If the wind generator were disconnected, what voltage would be supplied to
the lamp ?
(20 V)
(b)
At what wind speed would the generator begin to supply any current to the
lamp ?
(10 km/hr)
(c)
At what wind speed would the wind generator begin to charge the battery ?
(13.2 km/hr)
(d) As it stands, the circuit will suffer from a serious problem when the wind is light.
What is the problem and what can be done about it ?
Wind
Generator
2
ohm
+
-
2u
volt
4
ohm
Light .. resistance
20 ohm
Battery
+
24
- volt
8. A 12-V car battery supplies two 6-w rear light bulbs and two 30-w headlamp bulbs.
Sketch the circuit and calculate the total current drawn from the battery. What would
the current and battery terminal voltage be if the battery had an internal resistance of
0.5 ohm ? (6 A, 4.8 A, 9.6 V)