Nilsson/Riedel 7/e
12/23/2003
Selected Student Solutions for Chapter 2
Assessment Problems
2.5 First note that we know the current through all elements in the circuit except the 6
kW resistor (the current in the three elements to the left of the 6 kW resistor is i1; the
current in the three elements to the right of the 6 kW resistor is 30i1). To find the
current in the 6 kW resistor, write a KCL equation at the top node:
i1 + 30i1 = i6 k = 31i1
We can then use Ohm’s law to find the voltages across each resistor in terms of i1.
The results are shown in the figure below:
a) To find i1, write a KVL equation around the left-hand loop, summing voltages in a
clockwise direction starting below the 5V source:
- 5V + 54,000i1 - 1V + 186,000i1 = 0
Solving for i1,
54,000i1 + 186,000i1 = 6V
240,000i1 = 6V
6
= 25mA
240,000
b) Now that we have the value of i1, we can calculate the voltage for each
component except the dependent source. Then we can write a KVL equation for
the right-hand loop to find the voltage v of the dependent source. Sum the
voltages in the clockwise direction, starting to the left of the dependent source:
+ v - 54,000i1 + 8V - 186,000i1 = 0
v = 240,000i1 - 8V
i1 =
Thus, v = -2V.
= 240,000(25 ¥ 10 -6 ) - 8V
= 6V - 8V
= -2V
Nilsson/Riedel 7/e
12/23/2003
We now know the values of voltage and current for every circuit element. Let’s
construct a power table:
Element
5V
54kW
1V
6kW
Dep. Source
1.8kW
8V
Current
(mA)
25
25
25
775
775
775
775
Voltage
(V)
5
1.35
1
4.65
-2
1.35
8
Power
equation
p = -vi
p = Ri2
p = -vi
p = Ri2
p = -vi
p = Ri2
p = -vi
Power
(mW)
-125
33.75
-25
3603.75
1500
1012.5
-6000
c) The total power generated in the circuit is the sum of the negative power values in
the power table:
- 125mW + -25mW + -6000mW = -6150mW
Thus, the total power generated in the circuit is 6150 mW .
d) The total power absorbed in the circuit is the sum of the positive power values in
the power table:
33.75mW + 3603.75mW + 1500mW + 1012.5mW = 6150mW
Thus, the total power absorbed in the circuit is 6150 mW .
2.6 Given that if = 2A, we know the current in the dependent source is 2if = 4A. We can
write a KCL equation at the left node to find the current in the 10W resistor.
Summing the currents leaving the node,
- 5A + 2A + 4A + i10W = 0
i10W = 5A - 2A - 4A = -1A
Thus, the current in the 10W resistor is 1A, flowing right to left, as seen in the circuit
below.
a) To find vs, write a KVL equation, summing the voltages counter-clockwise
around the lower right loop. Start below the voltage source.
- v s + (1A)(10W) + (2A)(30W) = 0
- v s + 10V + 60V = 0
v s = 70V
Nilsson/Riedel 7/e
12/23/2003
b) The current in the voltage source can be found by writing a KCL equation at the
right-hand node. Sum the currents leaving the node
- 4A + 1A + i v = 0
iv = 4A - 1A = 3A
The current in the voltage source is 3A, flowing top to bottom. The power
associated with this source is
p = vi = (70V)(3A) = 210W
Thus, 210W are absorbed by the voltage source.
c) The voltage drop across the independent current source can be found by writing a
KVL equation around the left loop in a clockwise direction:
- v5A + ( 2A)(30W) = 0
- v 5A + 60V = 0
v 5A = 60V
The power associated with this source is
p = -v5A i = ( -60V)(5A) = -300W
This source thus delivers 300W of power to the circuit.
d) The voltage across the controlled current source can be found by writing a KVL
equation around the upper right loop in a clockwise direction:
+ v4 A + (10W)(1A) = 0
v 4A + 10V = 0
v 4A = -10V
The power associated with this source is
p = v4 A i = ( -10V)(4A) = -40W
This source thus delivers 40W of power to the circuit.
e) The total power dissipated by the resistors is given by
(i30W ) 2 (30W) + (i10W ) 2 (10W) = ( 2) 2 (30W) + (1) 2 (10W) = 120 + 10 = 130W
Chapter Problems
2.30 To find the power absorbed be each resistor, we find the current through each
resistor and use the equation p=i2R to calculate power. We label the voltages and
currents in the circuit as shown below.
Nilsson/Riedel 7/e
12/23/2003
We begin with the 22.5W resistor. Since we know its voltage, we can find its
current using Ohm’s law:
90V
= 4A
22.5W
Next, we can find v14 by writing a KVL equation around the outer loop of the
circuit. Start below the voltage source and sum the voltages in a clockwise
direction:
- 240V + 90V + v 15 = 0
i22.5 =
v15 = 240V - 90V = 150V
Next, find the current i15 using Ohm’s law:
150V
i15 =
= 10A
15W
Now find the current i5 by writing a KCL equation at the right-most node. Sum
the currents leaving this node.
i15 - i5 - i22.5 = 0
i5 = i15 - i22.5 = 10A - 4A = 6A
Then find the voltage v5 using Ohm’s law:
v5 = (5W)(6A) = 30V
Now write a KVL equation for the top loop to find the voltage v4. Sum the
voltages in the clockwise direction.
90V - v 5 - v4 = 0
v4 = 90V - v5 = 90V - 30V = 60V
Find the current i4 using Ohm’s law:
60V
i4 =
= 15A
4W
Now write a KVL equation for the bottom loop to find the voltage v20. Sum the
voltages in the clockwise direction.
v 15 - v 20 + v5 = 0
v20 = v15 - v5 = 150V + 30V = 180V
Finally, find the current i20 using Ohm’s law:
180V
i20 =
= 9A
20W
The voltages and currents found above are summarized in the circuit below.
Nilsson/Riedel 7/e
12/23/2003
a) We can find the power absorbed by each resistor using either p = i2R or p =
v2/R. Just to be consistent, we will use p = i2R:
p4 = (15) 2 ( 4) = 900W
p5 = (6) 2 (5) = 180W
p22.5 = ( 4) 2 ( 22.5) = 360W
p20 = (9) 2 ( 20) = 1620W
p15 = (10) 2 (15) = 1500W
b) To find the power delivered by the voltage source, we must find the current in
the voltage source, ig. We can find this current by writing a KCL equation at
the top node. Summing the currents leaving,
- i g + 15A + 4A = 0
i g = 15A + 4A = 19A
Find the power associated with the voltage source using p = -vi. (Note that the
current ig flows into the minus terminal of the voltage source).
p240V = -( 240)(19) = -4560W
Thus, the voltage source supplies 4560 W to the circuit.
c) The total power dissipated is the sum of the power values associated with all
of the resistors.
pdiss = 900W + 180W + 360W + 1620W + 1500W = 4560W
2.33 First note that we know the current through all elements in the circuit except the
200 W resistor (the current in the three elements to the left of the 200 W resistor is ib;
the current in the three elements to the right of the 200 W resistor is 49ib). To find
the current in the 200 W resistor, write a KCL equation at the top node:
i1 + 49ib = i200 = 50ib
We can then use Ohm’s law to find the voltages across each resistor in terms of ib.
The results are shown in the figure below:
a) Our approach is to find ib first, and then use the known value of ib to find vy. To
find i1, write a KVL equation around the left-hand loop, summing voltages in a
clockwise direction starting below the 7.2V source:
- 7.2V + 55,000i b + 0.7V + 10,000i b = 0
Solving for ib,
55,000i b + 10,000i b = 6.5V
65,000i b = 6.5V
ib =
6.5
= 100mA
65,000
Nilsson/Riedel 7/e
12/23/2003
Now that we have the value of ib, we can calculate the voltage for each
component except the dependent source. Then we can write a KVL equation for
the right-hand loop to find the voltage v of the dependent source. Sum the
voltages in the clockwise direction, starting to the left of the dependent source:
- vY - 24,500ib + 9V - 10,000i b = 0
vY = 9V - 34,500ib
Thus, vy = 5.55V.
= 9V - 34,500(100 ¥ 10 -6 )
= 9V - 3.45V
= 5.55V
b) We now know the values of voltage and current for every circuit element. Let’s
construct a power table:
Element
7.2V
55kW
0.7V
200W
Dep. Source
500W
9V
Current
(mA)
100
100
100
5000
4900
4900
4900
Voltage
(V)
7.2
5.5
0.7
1
5.55
2.45
9
Power
equation
p = -vi
p = Ri2
p = vi
p = Ri2
p = vi
p = Ri2
p = -vi
Power
(mW)
-720
550
70
5000
27,195
12,005
-44,100
The total power generated in the circuit is the sum of the negative power values in
the power table:
- 729 mW + -44,100mW = -44,820mW
Thus, the total power generated in the circuit is 44,820 mW .
The total power absorbed in the circuit is the sum of the positive power values in
the power table:
550mW + 70mW + 5000mW + 27,195mW + 12,005mW = 44,820mW
Thus, the total power absorbed in the circuit is 44,820, which equals the total
power generated.