ELEC166
Tutorial Week 2 Solutions
Q1. In the water flow model for electrical circuits, what quantity corresponds to the
current flowing in a wire?
A:
The current corresponds to the water flow rate − that is, the volume of water per
second passing through a pipe.
Q2. In the water flow model for electrical circuits, what quantity corresponds to the
voltage difference between two points in a circuit?
A:
The voltage difference corresponds to the difference in water pressure between
two points. One physical way of producing a pressure difference is to create a
height difference. Consider two tanks of water:
•
•
If we put both tanks on the table (same height), no water flows. In an
electrical circuit, this corresponds to no current flowing.
If the tanks are placed at different heights, water would flow from the higher
to the lower reservoir; the greater the difference in height, the greater would
be the flow and it happens the same way in electrical circuits. The height
difference can be considered as a potential difference in a circuit and current
flows from higher potential to lower potential like water flowing from the
higher tank to the lower one.
Q3. In the water flow model for electrical circuits, what corresponds to a resistor, and
how does the water model for a resistor correspond to its electrical properties
(say, for a length of wire used as a resistor)?
A:
A pipe in the water model corresponds to a resistor. The measure of how hard it
it for water to flow through a pipe corresponds to the electrical resistance of a
wire. Clearly it is easier for water to flow through a fat pipe than through a
narrow one, and the same is true for current flowing in a thick compared to a
thin conductor. In the water model, a long pipe will offer more resistance than a
short one, and the same applies for wires.
Q4. Does the amount of current change when it flows through a resistor? How does
this compare with the water flow model?
A:
Current does not change when it flows through a resistor, just like the water
flowing through a pipe will not appear or disappear (even though its velocity may
change). However, when branching via two pipes of unequal cross-sectional area,
more water will flow in the fat pipe than in the thin one. The same applies to
current. When current flows through two parallel branches, more current will
flow through the branch with lesser resistance than the one with larger
resistance.
ELEC166 Tutorial Week 2 Solutions
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Q5. A current (electron flow) carries energy. Where does the energy go when a
current flows through a resistor?
A:
Notice that current is not lost along the way, but the energy carried by the
electrons is. The energy goes into heating the resistor (we say that it is dissipated
in the resistor).
Q6. How many different ways can you connect together four 8Ω resistors, and what is
the resistance value of each of these combinations?
A:
There are in fact ten different ways to connect four resistors (apart from just
changing the order). Note that configurations E and F have the same overall
resistance.
A: 32Ω
G: 10.67Ω
B: 20Ω
H: 4.8Ω
C: 13.33Ω
D: 6Ω
I: 3.2Ω
E: 8Ω
F: 8Ω
J: 2Ω
Q7. Three resistors are connected in series across a 6V battery. If the resistors have
values 8Ω, 16Ω and 24Ω, what is the voltage across each resistor?
A:
The three resistors form a potential divider, so that the voltage across each
resistor is proportional to its resistance. So:
8Ω
= 1V
Voltage across 8Ω = 6V ×
8Ω + 16Ω + 24Ω
16Ω
Voltage across 16Ω = 6V ×
= 2V
8Ω + 16Ω + 24Ω
24Ω
Voltage across 24Ω = 6V ×
= 3V
8Ω + 16Ω + 24Ω
Alternatively, we can go back and use Ohm’s Law more directly. The total
resistance in the circuit is 8 + 16 + 24 = 48Ω, so the current (the same for each
resistor) is 6/48=0.125A. So for the 8Ω resistor, V = IR = 8Ω×0.125=1V, and
similarly for the other resistors.
Q8.
If three 360Ω resistors are connected in parallel across a 6V battery, calculate
the total current supplied by the battery.
ELEC166 Tutorial Week 2 Solutions
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−1
A:
Q9.
A:
1
1 ⎞
1 1
1
1
⎛ 1
+
+
, we have that R = ⎜
Using
= +
+
⎟ = 120Ω , and hence
R R1 R2 R3
⎝ 360 360 360 ⎠
V
6V
I= =
= 0.05 A = 50mA .
R 120Ω
When a current of 5mA flows through a resistor of 5kΩ, what is the voltage
across the resistor? What power will be dissipated by the resistor?
Using V=IR, we find that V=25V. Using P=VI or V2/R or I2R gives a power of
0.125W, or 125mW.
Q10. Draw a circuit diagram for each of the following:
a) Two resistors connected in parallel and then connected in series with a DC
voltage source.
b) Two resistors and a DC voltage source all joined in series.
c) Two resistors connected in parallel and then connected in parallel with a
series combination of three resistors. This arrangement is then connected in
series with a DC voltage source and a diode.
If the values of all the resistors are 1Ω and the voltage source is 5V, calculate
the current supplied by the voltage source. Assume here that the diode behaves
like a simple resistor of value 100Ω.
A:
Suitable answers are shown in the diagram below. Reducing the circuits to their
simplest form by using rules for resistors in series and parallel, we find that the
currents in the three circuits are 10A, 2.5A and 49.8mA respectively.
Q11. Match the following symbols with their names.
ELEC166 Tutorial Week 2 Solutions
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resistor
capacitor
switch
current source
DC voltage source
terminals
inductor
NO connection
battery
connection
diode
A:
Going down the left-hand then right-hand lists of pictures, the correct order is:
resistor, terminals, capacitor, diode, switch, battery, DC voltage source, inductor,
connection, no connection, current source.
Q12. Calculate the output voltage across the lower 10Ω resistor in the following
voltage divider circuit.
5Ω
15V
10Ω
10Ω
A:
Output
This is basically a voltage divider with resistors of 15Ω and 10Ω. By using the
equation for a potential divider, or other methods, the output voltage is 6V.
For example, you could proceed this way: the total resistance in the circuit is
5+10+10=25Ω, so the current flowing in each resistor will by Ohm’s Law (I=V/R)
be 15V/25Ω=0.6A. Now use Ohm’s Law again to get the voltage across the output
resistor: V = I×R = 0.6A × 10Ω = 6V.
Q13. Examine the following circuit diagram:
ELEC166 Tutorial Week 2 Solutions
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R1
R3
R2
A
R5
R6
R7
R8
R9
R10
A:
R4
R13
R11
R12
(a)
What would you do to measure the potential difference across R7?
(b)
Which resistors’ current in the circuit does the ammeter measure?
(c)
How would you modify the circuit so that you can measure the current
flowing through R1? Which resistor will also have the same amount of
current flowing through it? Why?
(d)
In a series of circuit diagrams, successively replace parts of the circuit by
their equivalent circuits, and reduce the whole circuit to a single voltage
source and resistor. Assume that the value of each resistor is 1kΩ and the
internal resistance of the ammeter is zero.
(a)
(b)
(c)
By connecting a voltmeter in parallel with R7.
The ammeter measures the current through R3 and R4.
To measure the current you would connect an ammeter in series with R1.
R2 will have the same current as R1 because it is in series with R1.
The final equivalent resistance of the circuit is approximately 3164Ω.
(d)
Q14. Show that, when N resistors, each having resistance R, are connected in
parallel, the total resistance is R/N.
A:
1
1
1
1
=
+
+ ... +
. For N equal resistors we
Rtot R1 R2
RN
N
1
1 1
1
R
have that
. Hence Rtot =
= + + ... +
=
.
Rtot R R
R
R
N
We can use the formula
(You might also try to show this using Ohm’s Law directly. For a given voltage
across the resistors, calculate the current in each resistor; sum the currents, then
use Ohm’s Law again to calculate the total effective resistance of the
combination.)
ELEC166 Tutorial Week 2 Solutions
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Q15. Here are three resistor networks:
1
A
1Ω
C 2Ω
3Ω
B
4Ω
2Ω
2
A
3Ω
B
1Ω
C
4Ω
5Ω
A
3
1Ω
C 2Ω
5Ω
3Ω
B
4Ω
6Ω
(a)
For each of the three networks, calculate the resistance as seen between
points A and B.
(b)
For each of the three networks, calculate the resistance as seen between
points C and B.
(c)
If instead points A and B were connected together (i.e. by a short circuit),
calculate the resistance as seen between points C and B for each of the
three networks.
(a)
The steps are:
Circuit 1: 2Ω in series with 3Ω (= 5Ω), result in parallel (||) with 4Ω
(=2.222Ω), result in series with 1Ω (=3.222Ω).
A:
Circuit 2: 2Ω in series with 3Ω (= 5Ω), result ||4Ω (=2.222Ω), result in
series with 1Ω (=3.222Ω), result ||5Ω (=1.959Ω).
Circuit 3: 1Ω||3Ω (=0.75Ω), in series with (2Ω||4Ω=1.333Ω) (=2.083Ω),
result in series with 5Ω (=7.083Ω), result ||6Ω (=3.248Ω)
(b)
Redraw the circuits as shown below. Notice that, in circuit 1, the 1Ω
resistor is not included, since the left-hand end of it is unconnected, and
thus it will have no effect on the circuit.
ELEC166 Tutorial Week 2 Solutions
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1
C
2Ω
3Ω
B
4Ω
2Ω
2
3Ω
C
1Ω
C
B
4Ω
A 5Ω
2Ω
5Ω
B
4Ω
1Ω
3
3Ω
A
6Ω
So the steps are:
Circuit 1: 2Ω in series with 3Ω (= 5Ω), result in parallel with (||) 4Ω
(=2.222Ω).
Circuit 2: 2Ω in series with 3Ω (= 5Ω), result ||4Ω (=2.222Ω), result ||
(6Ω=1Ω+5Ω) (=1.622Ω).
Circuit 3: The upper set of 3 resistors and the lower set of 3 resistors are in
parallel. The upper group’s resistance between A and B is (2Ω||4Ω) in
series with 5Ω (=6.333Ω). Similarly, the lower group has a total resistance
of (1Ω||3Ω)+6Ω = 6.75Ω. The total resistance is therefore 6.333Ω||6.75Ω
= 3.267Ω.
(c)
Redraw the circuits as follows. Notice that, in circuit 1, the1Ω resistor
must be included, since the left-hand end of it (point A) is connected to B.
The resistor thus forms part of the whole circuit.
1Ω
1
C
2Ω
3Ω
A,B
4Ω
2Ω
2
3Ω
C
4Ω
1Ω
C
3
A,B
A
2Ω
A,B
5Ω
4Ω
1Ω
3Ω
A
The final answers are then:
Circuit 1: 0.689Ω, Circuit 2: 0.689Ω, Circuit 3: 0.671Ω.
ELEC166 Tutorial Week 2 Solutions
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