Solutions
1. In the circuit shown, resistor C has a resistance R and the voltage across the battery is V . The power delivered to resistor C is 3 times as great as the power delivered to resistor A . The current through resistor C is one-fourth the current through resistor
B .
(A) Write expressions, in terms of R , for the resistances of resistors A and B . Explain your reasoning.
(B) Write an expression, in terms of V and R , for the current through resistor A .
(C) Suppose V = 12 V and R = 24 Ω. Find the total power dissipated by the circuit.
A
C
R
+
V
−
B
(A) Resistors A and C are in series, so they have the same current. We know P = i 2 R , so if the current is the same, then power is proportional to resistance. Therefore resistor
C has 3 times the resistance of A: R
A
= R/ 3
The voltage drop across resistor B is V . This is the same as the sum of voltage drops across resistors A and C. But the total resistance of A and C is
4 R/ 3. Since A and C carry 1/4 the current of B, we may write
R
AC
= R + R/ 3, or
V = iR
B
V = ( i/ 4)(4 R/ 3) = iR/ 3
Thus iR
B
= iR/ 3, or R
B
= R/ 3
(B) The current through A is just i
A
=
V
R
AC
V
=
4 R/ 3
= 3V/4R
(C) If V = 12 V and R = 24 Ω, then R
B
= 8 Ω and the power dissipated by B is
P
B
V 2
=
R
B
(12 V)
= =
= 18 W
8 Ω
2

Similarly, the power dissipated by A and C together is just
P
AC
V 2
=
R
AC
(12 V) 2
= =
4(24 Ω)/3
= 4.5 W
The total power is just 22.5 W .
Alternately, we could have found the total equivalent resistance, which is 6.4 Ω, and used P = V 2 /R to obtain the same result.
2. In the circuit below, find (A) the voltage across the 2- Ω resistor, and (B) the current supplied by the battery.
8 Ω
6 Ω
2 Ω
12 V
+
−
8 Ω
4 Ω 4 Ω 6 Ω
(A) First, we note that the leg containing the parallel 4- Ω and the 6- Ω resistors will not affect the answer: we could take it out and the voltage across the other parallel
“leg” would remain the same. So we may redraw the circuit:
8 Ω
8 Ω 8−V drop
12 V drop
8 Ω
2 Ω
6 Ω
4 Ω
4−V drop
In the figure, we combine the lower 8-ohm resistor with the 2- and 6- Ω resistors to form a 4- Ω resistor. It is clear that the voltage drop across the 4- Ω resistor is 4 volts.
This is also the drop across the series combination of 2- Ω and 6- Ω resistors. Therefore, treating the 2- Ω and 6- Ω resistors as a voltage divider, we find the drop across the
2- Ω resistor to be 1 volt .

(B) We are already part way there to finding the current. We need to find the total equivalent resistance. The equivalent resistance of the resistors in the last figure is 12 ohms. The equivalent resistance of the parallel 4- Ω and series 6- Ω resistors is 6 Ω +
2 Ω = 8 Ω. Therefore we have an 8 Ω resistor in parallel with a 12 Ωresistor.
R
1 eq
R eq
=
1
12Ω
+
1
8Ω
= 0.20833Ω
−
1
= 4.80 Ω
The battery current is then i =
V
R eq
=
12 V
4.80 Ω
= 2.5 A

3. Consider the circuit below. (A) Determine the current flowing through each resistor.
Series-parallel reduction does not work on this circuit. I suggest you use the mesh loop version of Kirchoff’s Rules to solve this network. (If you try to use the KVL/KCL method, you’ll end up with 6 simultaneous equations to solve.)
10
Ω
15
Ω
A
10
Ω
20
Ω
5
Ω
B
22.0 V
(B) Now, you can use your results above to find the equivalent resistance between points A and B (i.e., between the battery terminals.)
(C) Find the equivalent resistance by using a “∆-Y” conversion, and compare with the result in (B).
The mesh currents are shown here:
10
Ω
15
Ω
A
10
Ω i
1 i
3
20
Ω i
2
5
Ω
B
22.0 V
Applying Kirchoff’s voltage rule:
Loop 1:
−
10 i
1
+ 20( i
2
− i
1
) + 10( i
3
− i
1
) = 0
Loop 2:
−
40 i
1
+ 20 i
2
+ 10 i
3
= 0
−
15 i
2
+ 5( i
3 − i
2
) + 20( i
1 − i
2
) = 0
−
20 i
1
+ 40 i
2
−
5 i
3
= 0
Loop 3: 22 V + 10( i
1
− i
3
) + 5( i
2
− i
3
) = 0
10 i
1
+ 5 i
2
−
15 i
3
= 22 V

Note: loop 3 uses the lower, 10 Ω and 5 Ω resistors.
Dividing each equation by 5 V or 10 V and changing signs where appropriate, we obtain:
4 i
1 −
2 i
2 − i
3
= 0
4 i
1 −
8 i
2
+ i
3
= 0
2 i
1
+ i
2
−
3 i
3
=
−
4 .
4 V
(1)
(2)
(3)
This set of equations is not difficult to solve by Cramer’s Rule or by matrix algebra on a calculator. We can do it by substitution:
•
Subtract 2 from 1:
6 i
2
−
2 i
3
= 0 or i
3
= 3 i
2
(4)
•
Subtract twice 3 from 2:
−
10 i
2
+ 7 i
3
= 8 .
8 (5)
•
Substitute from 4:
−
10 i
2
+ 7(3 i
2
) = 8 .
8 i
2
= 8 .
8
11
A or 0.80 A (6)
•
Substitute into 4: i
3
= 3(0.80 A) = 2.4 A (7)
•
Substitute into 1:
0 = 4 i
1
−
2(0.80 A)
−
2.4 A i
1
= 1.0 A (8)
To find the “real” currents we can add the mesh currents where appropriate, coming up with the result shown below:
1.0 A 0.80 A
A
10
Ω
10
Ω
0.20 A
20
Ω
15
Ω
5
Ω
B
1.4 A 1.6 A
2.4 A
22.0 V

(B) The equivalent resistance is
V
R =
I
22 V
=
2.40 A
= 9.17 Ω or 9.2 Ω
(C) To use the ∆-Y method, we convert the circuit as follows:
R
2 15
Ω
R
1
10
Ω
A R
5
20
Ω
B
R
3 10
Ω
R
4
5
Ω
R
7
22.0 V
R
2
R
6
R
8
R
4
From a handout on the Delta-Y method, you can easily calculate the values for the equivalent “Y” circuit:
R
R
R
6
7
8
=
=
=
R
1
R
3
R
1
+ R
3
+ R
5
R
1
R
5
R
1
+ R
3
+ R
5
R
3
R
5
R
1
+ R
3
+ R
5
10 Ω)(10 Ω)
=
10 Ω+ 10 Ω+ 20 Ω
10 Ω)(20 Ω)
=
10 Ω+ 10 Ω+ 20 Ω
10 Ω)(20 Ω)
=
10 Ω+ 10 Ω+ 20 Ω
=
100 Ω
40
= 5 Ω
= 5 Ω
= 2.5 Ω
Putting those values into the bottom circuit, we have 20- Ω and 10- Ω resistors in parallel, and a 2.5- Ω resistor in series with that pair. The equivalent resistance is
R eq
1
= 2 .
5 Ω +
20 Ω
= 2 .
5 Ω + 6 .
67 Ω
+
1
10 Ω
−
1
= 9.17 Ωor 9.2 Ω as before.

4. Consider two circuits with terminals a and b as shown.
50
Ω a
20 mA
+ b
10 V
50
Ω a b
Suppose we investigate the behavior of these circuits by attaching an instrument to the terminals. The instrument consists of a variable voltage source in series with an ammeter. We may thereby hold the terminal a at some voltage relative to b , and observe the current flowing between the terminals. You can do this (in thought) by putting a battery between points a and b , remembering that an ideal battery has a zero internal resistance.
Your problem is to plot the current (coming out of terminal a ) as a function of the voltage ( a relative to b ). To start, calculate the current when the voltage is 0, +5 v,
+10 v, and -10 v. Plot these points and draw a straight line through them. Do this for both circuits.
Now, what is the interpretation of the slope of the line? Do you see why, referring to the Thevenin and Norton equivalents, that R
T H
= R
N OR
?
Both circuits result in the same plot, as shown:
I, amps
0.4
0.2
−10 −5 5
10
V, volts
−0.2
−0.4
The reciprocal of the slope is -50 Ω. This is the “internal” resistance of the circuit; the negative sign is only a result of the way we set up our voltage measurements.
In this case, 50 Ω is the change in voltage divided by the change in current; it expresses how much the current will change if you impose a certain voltage change on the Black
Box. This is what is meant by R
T H or R
N OR
, and it is clear that they are the same.