Brock University
Physics 1P22/1P92
Winter 2014
Dr. D’Agostino
Solutions for Tutorial 7: Circuits (Chapter 23)
1. (a) The circuit in the figure has two resistors, with R1 > R2 . Which resistor dissipates
the greater amount of power? Explain. [2 points]
Solution: In a series circuit the current is the same in each resistor. The power
dissipated in the first resistor is therefore P1 = I 2 R1 , and the power dissipated in
the second resistor is P2 = I 2 R2 . Because R1 > R2 , it follows that I 2 R1 > I 2 R2 ,
which means that P1 > P2 .
Thus, in a series circuit more power is dissipated in the resistor with greater
resistance.
(b) The circuit in the figure has two resistors, with R1 > R2 . Which resistor dissipates
the greater amount of power? Explain. [2 points]
Solution: In a simple parallel circuit, such as the one illustrated, the voltage
drop across each branch of the circuit is the same. The power dissipated in the
first resistor is P1 = V 2 /R1 , and the power dissipated in the second resistor is
P2 = V 2 /R2 .
Note that if R1 > R2 , then
1
1
<
R1
R2
It follows that
V2
V2
<
R1
R2
which means that P1 < P2 .
Thus, in a simple parallel circuit, more power is dissipated in the resistor with
smaller resistance.
(c) The five light bulbs in the figure have identical resistance. Rank the bulbs in
order of brightness, from brightest to dimmest. Explain. [6 points]
Solution: Let R represent the resistance of each individual light bulb, and consider
the circuit to consist of two parallel branches. One of the branches consists of light
bulbs A, B, and C, and the other consists of D and E. Clearly the parallel branch
containing A, B, and C is a bit complex, but nevertheless, it will be helpful to think
of the circuit in these terms.
The resistance of the ABC branch of the circuit is 1.5R and the resistance of the DE
branch of the circuit is 2R. (Verify this!) Thus, the current through resistor A is
IA =
2 V
V
= ·
1.5R
3 R
and the current through resistors D and E is
ID = IE =
1 V
V
= ·
2R
2 R
The current that passes through resistor A separates into two equal parts when it
passes through resistors B and C; that is,
IB = IC =
1 2 V
1 V
· · = ·
2 3 R
3 R
Thus,
IA > ID = IE > IB = IC
Because the power dissipated in each resistor is P = I 2 R, it follows that
PA > PD = PE > PB = PC
Thus, the order of brightness of the bulbs is
A>D=E>B=C
2. Determine the current leaving the battery, the current in each branch of the circuit,
and the power dissipated in each resistor. [10 points]
10 Ω
20 Ω
100 V
30 Ω
20 Ω
30 Ω
10 Ω
Solution: First determine the equivalent resistance of the circuit. You can do this by
replacing the circuit in turn with simpler and simpler ones having the same equivalent
resistance. The first step is to notice that the three resistors furthest to the right in the
circuit are connected in series, so they can be replaced by a single resistor of resistance
20 Ω + 30 Ω + 10 Ω = 60 Ω
The resulting simplified circuit is
10 Ω
100 V
30 Ω
60 Ω
20 Ω
Notice that the 30 Ω and 60 Ω resistors are in parallel in the previous diagram; thus,
they can be replaced by a single resistor of resistance
−1 −1 −1
1
1
2
1
3
60 Ω
+
=
+
=
=
= 20 Ω
30 Ω 60 Ω
60 Ω 60 Ω
60 Ω
3
The resulting simplified circuit is
10 Ω
100 V
20 Ω
20 Ω
We are left with a series circuit, and therefore the equivalent resistance is
10 Ω + 20 Ω + 20 Ω = 50 Ω
Because all of the circuits used in our chain of reasoning are equivalent, it follows that
the equivalent resistance of the original circuit is also 50 Ω. This is the resistance that
the battery “feels” or “sees” (depending on how you wish to anthropomorphize the
battery) as it tries to push current through the circuit, so (by Ohm’s law) the current
from the battery is
100 V
∆V
=
=2A
I=
R
50 Ω
Now let’s determine the current through each branch of the circuit. I’ll do this in
two ways; first, let’s redraw the circuit diagram and label key points in the circuit, in
preparation for determining the potential at the key points:
2
10 Ω
20 Ω
3
100 V
4
30 Ω
1
20 Ω
6
30 Ω
10 Ω
5
As usual, we are free to choose one point in the circuit and label it as the zero point
for potential; as usual, I’ll choose the “low end” of the battery as the zero point, so
V1 = 0. Then the “high end” of the battery has potential of V2 = 100 V.
Next let’s determine V3 and V6 . We can determine V3 by noting that all of the current
from the battery passes through the 10 Ω resistor between points 2 and 3 in the circuit,
so by Ohm’s law the potential drop between points 2 and 3 is
∆V = IR = (2 A)(10 Ω) = 20 V
Thus,
V3 = V2 − 20 V = 100 V − 20 V = 80 V
Similarly, we can determine V6 by noting that all of the current that flows into the
battery flows through the 20 Ω resistor between points 1 and 6 in the circuit. Therefore,
by Ohm’s law, the potential drop between points 1 and 6 is
∆V = IR = (2 A)(20 Ω) = 40 V
Thus,
V6 = V1 + 20 V = 0 V + 40 V = 40 V
Do you understand why we used addition in the previous equation and subtraction in
the calculation for V3 ? Think in terms of the playground analogy, and ask yourself
which direction is “downhill” when current goes through each resistor, and you will
clarify this.
Now that we know V3 and V6 , we can calculate the current through the 30 Ω resistor
between points 3 and 6 using Ohm’s law:
I1 =
V3 − V6
80 V − 40 V
40 V
4
∆V
=
=
=
= A
R
30 Ω
30 Ω
30 Ω
3
By Kirchhoff’s junction law, we can determine the current I2 in the right branch of the
circuit:
4
2
I = I1 + I2 ⇒ I2 = I − I1 = 2 A − A = A
3
3
We can now determine the power dissipated by each resistor, using P = I 2 R:
P23 = (2 A)2 (10 Ω) = 40 W
2
2
80
A (20 Ω) =
W = 8.9 W
P34 =
3
9
2
2
120
P45 =
A (30 Ω) =
W = 13.3 W
3
9
2
2
40
P56 =
A (10 Ω) =
W = 4.4 W
3
9
2
4
480
A (30 Ω) =
W = 53.3 W
P36 =
3
9
P16 = (2 A)2 (20 Ω) = 80 W
As a check on the power calculations, note that the battery “sees” a an equivalent
resistance in the circuit of 50Ω, so the batters “sees” a total power dissipation of
P = I 2 R = (2 A)2 (50 Ω) = 200 W
If you sum the individual power dissipations in each resistor, you’ll find that the total
is indeed 200 W (watch out for rounding errors).
A second solution to the problem involves an alternative method for determining the
currents in each branch of the circuit. You could begin with the following diagram:
10 Ω
20 Ω
I
I2
I1
100 V
20 Ω
30 Ω
30 Ω
10 Ω
Applying Kirchhoff’s junction law to the junction between the two upper resistors, we
obtain
I = I1 + I2
(1)
Applying Kirchoff’s loop law to the left loop (i.e., starting at point 1, then going
through points 2, 3, 6, and back to 1), we obtain
100 − 10I − 30I1 − 20I = 0
(2)
Applying Kirchoff’s loop law to the right loop (i.e., starting at point 6, then going
through points 3, 4, 5, and back to 6), we obtain
30I1 − 20I2 − 30I2 − 10I2 = 0
(3)
Equations (1), (2), and (3) are three independent equations for the three unknowns I,
I1 , and I2 , and so you can use them to determine the three currents without further
reference to the diagram. Once you have the currents, the rest of the solution (to obtain
the power dissipated in each resistor) proceeds exactly as in the previous solution.
3. Determine the current leaving each battery and the power dissipated in each resistor.
[10 points]
10 Ω
100 V
10 Ω
50 V
20 Ω
10 Ω
Solution: As in the previous problem, one can proceed by determining potentials
at key points in the circuit, or one can write three equations for the three unknown
currents using Kirchhoff’s laws. Here is what you might obtain using Kirchhoff’s laws:
10 Ω
10 Ω
I1
I3
100 V
50 V
I2
20 Ω
10 Ω
I1 + I2 = I3
100 − 10I1 − 50 = 0
50 − 10I3 − 20I3 − 10I3 = 0
Solve the three equations and you will obtain
I1 = 5 A
I2 = −3.75 A
I3 = 1.25 A
The fact that the value of I2 is negative means that our guess (in the diagram) for
the direction of this current is actually incorrect; in fact, the current I2 flows in the
direction opposite the arrow.
To complete the solution, you can determine that the power dissipated in each resistor
is as follows:
P1 = (5 A)2 (10 Ω) = 250 W
P2 = (1.25 A)2 (10 Ω) = 15.625 W
P3 = (1.25 A)2 (20 Ω) = 31.25 W
P4 = (1.25 A)2 (10 Ω) = 15.625 W
The total power dissipated in the circuit (which was not asked about) is therefore 312.5
W. Note that the battery on the left is sending energy into the circuit at the rate of
500 W. Of the 500 W, 312.5 W is dissipated in the resistors, which means that the
remaining power (187.5 W) serves to charge the battery on the right (50 V × 3.75 A
= 187.5 W).