ELEC166
Tutorial Week 3 Solutions
Q1. An ideal voltmeter gives a reading of 9V when measuring between the terminals
of a (real) battery. An ideal ohmmeter gives a reading of 900Ω when measuring
between the ends of a resistor. If the resistor is connected across the battery, will
a current of 10mA flow through the resistor, and, if not, why not?
A:
The current will be slightly lower than 10mA. This is due to the total resistance
in the circuit being larger than 900Ω when the internal resistance of the battery
is connected in series with the 900Ω resistor.
(Note that the 9V measured originally at the battery terminals was when there
was no current flowing. Ohm’s Law tells us that there was no voltage across the
battery’s internal resistance, so the 9V is an accurate measure of the only voltage
source in the circuit − that is, the Thévenin equivalent voltage.)
Q2. When a 10W bulb is connected in parallel with a 20W bulb, the 20W bulb is
brighter. Which bulb is brighter if they are connected in series? Explain.
A:
Note that in all our calculations with light bulbs, we assume that the resistance
of bulbs is the same whether they are lit or not; this is not correct, but simplifies
the calculations.
When the bulbs are connected in parallel, the same voltage is across each (this is
the normal way that they would be connected).
First find the relative resistances of the two bulbs; we don’t need to know the
V2
V2
voltage across them. Using P =
, we have that R20W =
for the 20W bulb, and
R
20
V2
R10W =
for the 10W bulb. Hence R10W=2×R20W.
10
There are two ways you might reach the final conclusion:
(1) (Simplest method) Since they are connected in series, the same current flows
through each bulb. Power is related to current and resistance by P=I2R, so for the
same current, the power is proportional to resistance. Since the resistance of the
10W bulb is higher, it dissipates more power (and therefore, presumably, is
brighter).
(2) Alternatively, consider the two bulbs in series forming a potential divider. If
the total voltage across the two bulbs is VS, then the voltage across the 10W bulb
2 × R20W
R10W
= VS ×
= 0.667VS , and the power dissipated is
is V10W = VS ×
2 × R20W + R20W
R10W + R20W
V 2 0.6672
0.4445 0.222
=
=
=
.
R
R10W
R20W
2 × R20W
Similarly, the voltage across the 20W bulb turns out to be 0.333VS, and the
V 2 0.3332 0.111
=
=
power dissipated is P =
. Thus the 20W bulb dissipates only
R
R20W
R20W
P=
ELEC166 Tutorial Week 3 Solutions
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half as much power as the 10W bulb; the 10W bulb is brighter.
Q3.
A:
Q4.
Given a 240V battery and four resistors of values 1Ω, 2Ω, 4Ω and 5Ω, design a
voltage divider circuit to produce an output voltage of 24V (you may not need to
use all the resistors).
We want an output which is one-tenth of the input voltage. Notice that
1+4+5=10, so by creating a voltage divider out of the 1Ω, 4Ω and 5Ω resistors the
1
voltage across the 1Ω resistor will be 240 ×
= 24V .
1+4 +5
Find the Thévenin equivalent of the following circuit, as seen between
terminals A and B. (HINT: Determine the open-circuit voltage and the shortcircuit current.)
3Ω
5V
A
Output
10V
B
5Ω
A:
Adding up the voltage sources around the circuit, we see that the open-circuit
voltage VOC is 10+5=15V. The short-circuit current ISC will thus be
15/(3+5)=1.875A. Following the procedure in the lecture notes, we find that the
Thévenin voltage is 15V and the Thévenin resistance (=VOC/ISC) is 8Ω.
(Alternatively, note that you can find the Thévenin resistance by simply looking
back into the terminals and replacing all the voltage sources with short circuits.)
Q5.
Find the Thévenin equivalent of the following circuit, as seen between
terminals A and B.
1Ω
5V
A
5V
Output
5Ω
B
10Ω
A:
This is the same as the previous question, except we now have two resistors in
parallel which we must reduce to their equivalent resistance before going
through the same steps as before. The Thévenin resistance is 4.33Ω and the
Thévenin voltage is 10V.
ELEC166 Tutorial Week 3 Solutions
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Q6.
For the following circuit, calculate (i) the voltage between terminals B and C,
and (ii) the voltage between terminals A and B. What do you notice about these
two voltages?
A
15V
1kΩ
2kΩ
B
5kΩ
3kΩ
C
A:
Q7.
This circuit forms a two-resistor potential divider, with each resistor being a
parallel combination of two others. The upper combination is
(1kΩ||2kΩ) = 0.667kΩ, while the lower combination is (3kΩ||5kΩ) = 1.875kΩ.
1.875kΩ
Thus the voltage between B and C is VBC = 15V ×
= 11.064V .
1.875kΩ + 0.667kΩ
The voltage between A and B is calculated in the same way, except that we now
want the voltage across the upper resistor pair.
0.667kΩ
This is VAB = 15V ×
= 3.936V .
1.875kΩ + 0.667kΩ
The two voltages must add up to 15V.
For the following circuit, algebraically calculate its Thévenin equivalent circuit
as seen between terminals A and B (do this in the standard way by calculating
the open-circuit voltage, short-circuit current and so on). The equivalent
resistance turns out to be a simple combination of R1 and R2 − what is it?
R1
Vs
A
R2
Vout
B
A:
Since this is a potential divider, the open-circuit voltage is given by
R2
VOC = VS ×
. This is the Thévenin equivalent voltage.
R1 + R2
If A and B are shorted together, all the circuit current will flow through the short
circuit, and none through R2, and since the total resistance in the circuit is just
V
R1, the short-circuit curent is given by I SC = S . The Thévenin equivalent
R1
VOC
R2
R1 R1 × R2
resistance is given by the ratio
. This is an
= VS ×
×
=
I SC
R1 + R2 VS R1 + R2
alternative form of the formula for the equivalent resistance of two resistors in
parallel. So the Thévenin equivalent resistance of a two-resistor potential divider
is equal to the parallel combination of the two resistors.
ELEC166 Tutorial Week 3 Solutions
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Q8.
Find the Thévenin equivalent of the following circuit, as seen between
terminals A and B.
12kΩ
15kΩ
A
Vs=12V
+
20kΩ
_
B
A:
First let’s calculate the open-circuit voltage VOC. If we ignore the 15kΩ resistor
for the moment, we see that the two left-hand resistors form a potential divider,
20kΩ
so that the voltage across the 20kΩ is V20kΩ = 12V ×
= 7.5V . Since
20kΩ + 12kΩ
the terminals A and B are open-circuited, no current flows through the 15kΩ, so
there is zero voltage across it. Hence the voltage between A and B is also 7.5V.
This is VOC, and hence the Thévenin equivalent voltage.
Now for the short-circuit current ISC. This will be equal to the current flowing in
the 15kΩ when A and B are connected together, as shown in the circuit below.
12kΩ
Vs=12V
+
_
20kΩ
15kΩ
Isc
In this case, the total resistance in the circuit will be
12kΩ+(15kΩ||20kΩ)=8.571kΩ. Hence the total current in the circuit will be
12V/8.571kΩ=1.400mA. However, this is not the current we want, since it is
shared between the 15kΩ and 20kΩ, but we can find ISC by calculating the
voltage across the 15kΩ.
Notice that the second circuit forms a potential divider. The voltage across the
15kΩ (and the 20kΩ) is thus given by
(20kΩ||15kΩ )
8.571kΩ
V15kΩ = 12V ×
= 12V ×
= 5.00V . Thus the current
12kΩ + (20kΩ||15kΩ )
20.571kΩ
through the 15kΩ is just 5V/15kΩ=0.333mA, which is ISC.
Now the Thévenin equivalent resistance is given by the ratio
VOC
7.5V
=
= 22.5kΩ .
I SC 0.333mA
ELEC166 Tutorial Week 3 Solutions
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Q9. Consider the circuit below. By initially converting blocks A and B into their
Thévenin equivalent circuits, calculate the current flowing in the 10kΩ resistor.
A
B
2kΩ
6kΩ
10kΩ
5V
10V
4kΩ
3kΩ
A:
Blocks A and B are simply potential dividers. We can use the result from a
previous question to quickly calculate the Thévenin equivalent circuits of these
blocks (recall that the value of the equivalent series resistor is equal to the
parallel combination of the two original resistors in the divider). So the overall
circuit can be reduced to the following:
10kΩ
1.2kΩ
2.4kΩ
+
+
_
_
3V
4V
All the components except the 10kΩ resistor can then be combined into another
Thévenin equivalent circuit (see another earlier question), so that the final
circuit can be redrawn like this (notice also that the two voltage sources oppose
each other, so that the resulting voltage is 4V−3V=1V):
3.6kΩ
_
10kΩ
+
1V
It is then a relatively simple matter to see that the current in the 10kΩ is equal
to 1V/13.6kΩ = 73.5µA.
Q10.
Instead of its Thévenin equivalent, a linear circuit can also be replaced by its
Norton equivalent circuit, which consists of a current source in parallel with a
resistor, as shown below. For a given circuit, if the Thévenin voltage source and
series resistor have values Vth and Rth respectively, calculate the values of the
corresponding Norton equivalent current source IN and parallel resistor RN in
terms of Vth and Rth. (Hint: As for the Thévenin case, consider the open-circuit
voltage and short-circuit current; they should be the same for both equivalent
circuits.)
ELEC166 Tutorial Week 3 Solutions
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Rth
Vth
⇔
+
IN
RN
_
Thévenin equivalent circuit
Norton equivalent circuit
A:
For the Thévenin equivalent circuit, the open-circuit voltage and shortcircuit current are equal to Vth and Vth/Rth respectively, while for the Norton
equivalent circuit they are equal to IN×RN and IN respectively (why?).
Equating these gives IN = Vth/Rth and RN = Rth.
ELEC166 Tutorial Week 3 Solutions
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