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Circuit Elements
Assessment Problems a
AP 2.
[a] Note that the current ib is in the same circuit branch as the 8 A current source; however, ib is defined in the opposite direction of the current source. Therefore, ib = −8 A Next, note that the dependent voltage source and the independent voltage source are in parallel with the same polarity. Therefore, their voltages are equal, and
vg =
ib 4
= −2 V
[b] To find the power associated with the 8 A source, we need to find the voltage drop across the source, vi. Note that the two independent sources are in parallel, and that the voltages vg and v 1 have the same polarities, so these voltages are equal:
vi = vg = −2 V Using the passive sign convention, ps = (8 A)(vi) = (8 A)(−2 V) = −16 W Thus the current source generated 16 W of power.
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2–2 CHAPTER 2. Circuit Elements
AP 2.
[a] Note from the circuit that vx = −25 V. To find α note that the two current sources are in the same branch of the circuit but their currents flow in opposite directions. Therefore
αvx = −15 A
Solve the above equation for α and substitute for vx,
α =
−15 A
vx
−15 A
−25 V
= 0.6 A/V
[b] To find the power associated with the voltage source we need to know the current, iv. Note that this current is in the same branch of the circuit as the dependent current source and these two currents flow in the same direction. Therefore, the current iv is the same as the current of the dependent source:
iv = αvx = (0.6)(−25) = −15 A
Using the passive sign convention,
ps = −(iv)(25 V) = −(−15 A)(25 V) = 375 W.
Thus the voltage source dissipates 375 W.
AP 2.
[a] The resistor and the voltage source are in parallel and the resistor voltage and the voltage source have the same polarities. Therefore these two voltages are the same:
vR = vg = 1 kV
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2–4 CHAPTER 2. Circuit Elements
conductance are in the same branch of the circuit so must have the same current. The voltage drop across the current source is vg, positive at the top, because the current source and the conductance are also in parallel so must have the same voltage. From a version of Ohm’s law,
vg =
ig G
0 .5 A
50 mS
= 10 V
Now that we know the voltage drop across the current source, we can find the power delivered by this source: psource = −vgig = −(10)(0.5) = −5 W
Thus the current source delivers 5 W to the circuit. [b] We can find the value of the conductance using the power, and the value of the current using Ohm’s law and the conductance value:
pg = Gv g^2 so G =
pg v^2 g
= 0.04 S = 40 mS
ig = Gvg = (40 mS)(15 V) = 0.6 A
[c] We can find the voltage from the power and the conductance, and then use the voltage value in Ohm’s law to find the current:
pg = Gv g^2 so v^2 g =
pg G
8 W
200 μS
Thus vg =
√ 40 ,000 = 200 V
ig = Gvg = (200 μS)(200 V) = 0.04 A = 40 mA
AP 2.5 [a] Redraw the circuit with all of the voltages and currents labeled for every circuit element.
Write a KVL equation clockwise around the circuit, starting below the voltage source:
−24 V + v 2 + v 5 − v 1 = 0
Next, use Ohm’s law to calculate the three unknown voltages from the three currents:
v 2 = 3i 2 ; v 5 = 7i 5 ; v 1 = 2i 1
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Problems 2–
A KCL equation at the upper right node gives i 2 = i 5 ; a KCL equation at the bottom right node gives i 5 = −i 1 ; a KCL equation at the upper left node gives is = −i 2. Now replace the currents i 1 and i 2 in the Ohm’s law equations with i 5 :
v 2 = 3i 2 = 3i 5 ; v 5 = 7i 5 ; v 1 = 2i 1 = − 2 i 5
Now substitute these expressions for the three voltages into the first equation:
24 = v 2 + v 5 − v 1 = 3i 5 + 7i 5 − (− 2 i 5 ) = 12i 5
Therefore i 5 = 24/12 = 2 A [b] v 1 = − 2 i 5 = −2(2) = −4 V [c] v 2 = 3i 5 = 3(2) = 6 V [d] v 5 = 7i 5 = 7(2) = 14 V [e] A KCL equation at the lower left node gives is = i 1. Since i 1 = −i 5 , is = −2 A. We can now compute the power associated with the voltage source:
p 24 = (24)is = (24)(−2) = −48 W
Therefore 24 V source is delivering 48 W.
AP 2.6 Redraw the circuit labeling all voltages and currents:
We can find the value of the unknown resistor if we can find the value of its voltage and its current. To start, write a KVL equation clockwise around the right loop, starting below the 24 Ω resistor:
−120 V + v 3 = 0
Use Ohm’s law to calculate the voltage across the 8 Ω resistor in terms of its current:
v 3 = 8i 3
Substitute the expression for v 3 into the first equation:
−120 V + 8i 3 = 0 so i 3 =
= 15 A
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Problems 2–
[b] Draw the circuit model from part (a) and attach a 25 Ω resistor:
To find the power delivered to the 25 Ω resistor we must calculate the current through the 25 Ω resistor. Do this by first using KCL to recognize that the current in each of the components is it, flowing in a clockwise direction. Write a KVL equation in the clockwise direction, starting below the voltage source, and using Ohm’s law to express the voltage drop across the resistors in the direction of the current it flowing through the resistors:
−25 V + 100it + 25it = 0 so 125 it = 25 so it =
= 0.2 A
Thus, the power delivered to the 25 Ω resistor is
p 25 = (25)i^2 t = (25)(0.2)^2 = 1 W.
AP 2.8 [a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0, it = 0.25 A. Therefore the current source must be 0.25 A. Since the plot is a straight line, its slope can be used to calculate the value of resistance:
R =
∆v ∆i
A circuit model having the same v − i characteristic is a 0.25 A current source in parallel with a 100Ω resistor, as shown below:
[b] Draw the circuit model from part (a) and attach a 25 Ω resistor:
Note that by writing a KVL equation around the right loop we see that the voltage drop across both resistors is vt. Write a KCL equation at the top center node, summing the currents leaving the node. Use Ohm’s law
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2–8 CHAPTER 2. Circuit Elements
to specify the currents through the resistors in terms of the voltage drop across the resistors and the value of the resistors.
− 0 .25 +
vt 100
vt 25
= 0, so 5 vt = 25, thus vt = 5 V
p 25 =
v t^2 25
= 1 W.
AP 2.9 First note that we know the current through all elements in the circuit except the 6 kΩ resistor (the current in the three elements to the left of the 6 kΩ resistor is i 1 ; the current in the three elements to the right of the 6 kΩ resistor is 30i 1 ). To find the current in the 6 kΩ resistor, write a KCL equation at the top node:
i 1 + 30i 1 = i6k = 31i 1
We can then use Ohm’s law to find the voltages across each resistor in terms of i 1. The results are shown in the figure below:
[a] To find i 1 , write a KVL equation around the left-hand loop, summing voltages in a clockwise direction starting below the 5V source:
−5 V + 54, 000 i 1 − 1 V + 186, 000 i 1 = 0 Solving for i 1
54 , 000 i 1 + 186, 000 i 1 = 6 V so 240 , 000 i 1 = 6 V
Thus,
i 1 =
= 25 μA
[b] Now that we have the value of i 1 , 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 , 000 i 1 + 8 V − 186 , 000 i 1 = 0
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2–10 CHAPTER 2. Circuit Elements
[a] To find vs, write a KVL equation, summing the voltages counter-clockwise around the lower right loop. Start below the voltage source. −vs + (1 A)(10 Ω) + (2 A)(30 Ω) = 0 so vs = 10 V + 60 V = 70 V
[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
−4 A + 1 A + iv = 0 so iv = 4 A − 1 A = 3 A
The current in the voltage source is 3 A, flowing top to bottom. The power associated with this source is
p = vi = (70 V)(3 A) = 210 W
Thus, 210 W 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:
−v 5 A + (2 A)(30 Ω) = 0 so v 5 A = 60 V
The power associated with this source is
p = −v 5 Ai = −(60 V)(5 A) = −300 W
This source thus delivers 300 W 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:
+v 4 A + (10 Ω)(1 A) = 0 so v 4 A = −10 V
The power associated with this source is
p = v 4 Ai = (−10 V)(4 A) = −40 W
This source thus delivers 40 W of power to the circuit. [e] The total power dissipated by the resistors is given by
(i30Ω)^2 (30 Ω) + (i10Ω)^2 (10 Ω) = (2)^2 (30 Ω) + (1)^2 (10 Ω) = 120 + 10 = 130 W
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Problems 2–
Problems
P 2.1 [a] Yes, independent voltage sources can carry the 5 A current required by the connection; independent current source can support any voltage required by the connection, in this case 5 V, positive at the bottom.
[b] 20 V source: absorbing 15 V source: developing (delivering)
5 A source: developing (delivering)
[c] P20V = (20)(5) = 100 W (abs) P15V = −(15)(5) = −75 W (dev/del)
P5A = −(5)(5) = −25 W (dev/del) ∑ Pabs =
∑ Pdel = 100 W [d] The interconnection is valid, but in this circuit the voltage drop across the 5 A current source is 35 V, positive at the top; 20 V source is developing (delivering), the 15 V source is developing (delivering), and the 5 A source is absorbing: P20V = −(20)(5) = −100 W (dev/del) P15V = −(15)(5) = −75 W (dev/del)
P5A = (35)(5) = 175 W (abs) ∑ Pabs =
∑ Pdel = 175 W
P 2.2 The interconnect is valid since the voltage sources can all carry 5 A of current supplied by the current source, and the current source can carry the voltage drop required by the interconnection. Note that the branch containing the 10 V, 40 V, and 5 A sources must have the same voltage drop as the branch containing the 50 V source, so the 5 A current source must have a voltage drop of 20 V, positive at the right. The voltages and currents are summarize in the circuit below:
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Problems 2–
Hence any combination of v 1 and v 2 such that v 1 + v 2 = −4 V is a valid solution.
P 2.6 [a] Because both current sources are in the same branch of the circuit, their values must be the same. Therefore, v 1 50
= 0. 4 → v 1 = 0.4(50) = 20 V
[b] p = v 1 (0.4) = (20)(0.4) = 8 W (absorbed)
P 2.7 [a] The voltage drop from the top node to the bottom node in this circuit must be the same for every path from the top to the bottom. Therefore, the voltages of the two voltage sources are equal:
−αi∆ = 6
Also, the current i∆ is in the same branch as the 15 mA current source, but in the opposite direction, so
i∆ = − 0. 015
Substituting,
−α(− 0 .015) = 6 → α =
The interconnection is valid if α = 400 V/A. [b] The voltage across the current source must equal the voltage across the 6 V source, since both are connected between the top and bottom nodes. Using the passive sign convention,
p = vi = (6)(0.015) = 0.09 = 90 mW
[c] Since the power is positive, the current source is absorbing power.
P 2.8 [a] Yes, each of the voltage sources can carry the current required by the interconnection, and each of the current sources can carry the voltage drop required by the interconnection. (Note that i 1 = 50 mA.) [b] No, because the voltage drop between the top terminal and the bottom terminal cannot be determined. For example, define v 1 , v 2 , and v 3 as shown:
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2–14 CHAPTER 2. Circuit Elements
The voltage drop across the left branch, the center branch, and the right branch must be the same, since these branches are connected at the same two terminals. This requires that v 1 − 20 = v 2 = v 3 + 30 But this equation has three unknown voltages, so the individual voltages cannot be determined, and thus the power of the sources cannot be determined.
P 2.9 The interconnection is invalid. In the middle branch, the value of the current ix must be 50 mA, since the 50 mA current source supplies current in this branch in the same direction as the current ix. Therefore, the voltage supplied by the dependent voltage source in the right hand branch is 1800(0.05) = 90 V. This gives a voltage drop from the top terminal to the bottom terminal in the right hand branch of 90 + 60 = 150 V. But the voltage drop between these same terminals in the left hand branch is 30 V, due to the voltage source in that branch. Therefore, the interconnection is invalid.
P 2.
First, 10va = 5 V, so va = 0.5 V. Then recognize that each of the three branches is connected between the same two nodes, so each of these branches must have the same voltage drop. The voltage drop across the middle branch is 5 V, and since va = 0.5 V, vg = 0. 5 − 5 = − 4 .5 V. Also, the voltage drop across the left branch is 5 V, so 20 + v9A = 5 V, and v9A = −15 V, where v9A is positive at the top. Note that the current through the 20 V source must be 9 A, flowing from top to bottom, and the current through the vg is 6 A flowing from top to bottom. Let’s find the power associated with the left and middle branches: p9A = (9)(−15) = −135 W p20V = (9)(20) = 180 W pvg = −(6)(− 4 .5) = 27 W p6A = (6)(0.5) = 3 W Since there is only one component left, we can find the total power: ptotal = −135 + 180 + 27 + 3 + pds = 75 + pds = 0 so pds must equal −75 W. Therefore, ∑ Pdev =
∑ Pabs = 210 W
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2–16 CHAPTER 2. Circuit Elements
P 2.14 Since we know the device is a resistor, we can use Ohm’s law to calculate the resistance. From Fig. P2.14(a),
v = Ri so R =
v i Using the values in the table of Fig. P2.14(b),
R =
= 1.2 kΩ
Note that this value is found in Appendix H.
P 2.15 Since we know the device is a resistor, we can use the power equation. From Fig. P2.15(a),
p = vi = v^2 R
so R = v^2 p
Using the values in the table of Fig. P2.13(b)
R =
(−8)^2
640 × 10 −^3
(−4)^2
160 × 10 −^3
(4)^2
160 × 10 −^3
(8)^2
640 × 10 −^3
(12)^2
1440 × 10 −^3
(16)^2
2560 × 10 −^3
Note that this value is found in Appendix H.
P 2.16 The resistor value is the ratio of the power to the square of the current: R =
p i^2
. Using the values for power and current in Fig. P2.16(b),
8. 25 × 10 −^3
(0. 5 × 10 −^3 )^2
33 × 10 −^3
(1 × 10 −^3 )^2
74. 25 × 10 −^3
(1. 5 × 10 −^3 )^2
132 × 10 −^3
(2 × 10 −^3 )^2
206. 25 × 10 −^3
(2. 5 × 10 −^3 )^2
297 × 10 −^3
(3 × 10 −^3 )^2
= 33 kΩ
Note that this is a value from Appendix H.
P 2.17 Label the unknown resistor currents and voltages:
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Problems 2–
[a] KCL at the top node: 0 .02 = i 1 + i 2 KVL around the right loop: −vo + v 2 − 5 = 0 Use Ohm’s law to write the resistor voltages in the previous equation in terms of the resistor currents:
− 5000 i 1 + 2000i 2 − 5 = 0 → − 5000 i 1 + 2000i 2 = 5
Multiply the KCL equation by −2000 and add it to the KVL equation to eliminate i 2 :
−2000(i 1 + i 2 ) + (− 5000 i 1 + 2000i 2 ) = −2000(0.02) + 5 → − 7000 i 1 = − 35
Solving,
i 1 =
= 0.005 = 5 mA
Therefore,
vo = Ri 1 = (5000)(0.005) = 25 V
[b] p20mA = −(0.02)vo = −(0.02)(25) = − 0 .5 W
i 2 = 0. 02 − i 1 = 0. 02 − 0 .005 = 0.015 A
p5V = −(5)i 2 = −(5)(0.015) = − 0 .075 W
p5k = 5000i^21 = 5000(0.005)^2 = 0.125 W
p2k = 2000i^22 = 2000(0.015)^2 = 0.45 W
ptotal = p20mA + p5V + p5k + p2k = − 0. 5 − 0 .075 + 0.125 + 0.45 = 0
Thus the power in the circuit balances.
P 2.18 [a]
20 ia = 80 ib ig = ia + ib = 5ib
ia = 4 ib
50 = 4 ig + 80ib = 20ib + 80ib = 100ib ib = 0 .5 A, therefore, ia = 2 A and ig = 2.5 A [b] ib = 0.5 A [c] vo = 80ib = 40 V
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Problems 2–
P 2.20 Label the unknown resistor voltages and currents:
[a] ia =
= 0.02 A (Ohm’s law)
i 1 = ia = 0.02 A (KCL)
[b] vb = 200i 1 = 200(0.02) = 4 V (Ohm’s law)
−v 1 + vb + 3.5 = 0 so v 1 = 3.5 + vb = 3.5 + 4 = 7.5 V (KVL)
[c] va = 0.05(50) = 2.5 V (Ohm’s law)
−vg + va + v 1 = 0 so vg = va + v 1 = 2.5 + 7.5 = 10 V (KVL)
[d] pg = vg(0.05) = 10(0.05) = 0.5 W
P 2.21 [a] Use KVL for the right loop to calculate the voltage drop across the right-hand branch vo. This is also the voltage drop across the middle branch, so once vo is known, use Ohm’s law to calculate io: vo = 1000 ia + 4000ia + 3000ia = 8000ia = 8000(0.002) = 16 V
16 = 2000 io
io =
= 8 mA
[b] KCL at the top node: ig = ia + io = 0.002 + 0.008 = 0.010 A = 10 mA. [c] The voltage drop across the source is v 0 , seen by writing a KVL equation for the left loop. Thus, pg = −voig = −(16)(0.01) = − 0 .160 W = −160 mW. Thus the source delivers 160 mW.
P 2.22 [a]
v 2 = 150 − 50(1) = 100V
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2–20 CHAPTER 2. Circuit Elements
i 2 =
v 2 25
= 4A
i 3 + 1 = i 2 , i 3 = 4 − 1 = 3A
v 1 = 10i 3 + 25i 2 = 10(3) + 25(4) = 130V
i 1 =
v 1 65
= 2A
Note also that
i 4 = i 1 + i 3 = 2 + 3 = 5 A
ig = i 4 + io = 5 + 1 = 6 A
[b] p4Ω = 52 (4) = 100 W
p50Ω = 12 (50) = 50 W p65Ω = 22 (65) = 260 W
p10Ω = 32 (10) = 90 W
p25Ω = 42 (25) = 400 W
[c]
∑ Pdis = 100 + 50 + 260 + 90 + 400 = 900 W
Pdev = 150ig = 150(6) = 900 W
P 2.23 Label all unknown resistor voltages and currents:
Ohms’ law for 5 kΩ resistor: v 1 = (0.01)(5000) = 50 V KVL for lower left loop: −80 + v 2 + 50 = 0 → v 2 = 80 − 50 = 30 V Ohm’s law for 1.5 kΩ resistor: i 2 = v 2 /1500 = 30/1500 = 20 mA KCL at center node: i 2 = i 3 + 0. 01 → i 3 = i 2 − 0 .01 = 0. 02 − 0 .01 = 0.01 = 10 mA Ohm’s law for 3 kΩ resistor v 3 = 3000i 3 = 3000(0.01) = 30 V KVL for lower right loop: −v 1 + v 3 + v 4 = 0 → v 4 = v 1 − v 3 = 50 − 30 = 20 V
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