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Circuit Elements
Assessment Problems
AP 2.
[a] To find vg write a KVL equation clockwise around the left loop, starting below the dependent source:
−
ib 4
ib 4 To find ib write a KCL equation at the upper right node. Sum the currents leaving the node:
ib + 8 A = 0 so ib = − 8 A
Thus,
vg =
= − 2 V
[b] To find the power associated with the 8 A source, we need to find the voltage drop across the source, vi. To do this, write a KVL equation clockwise around the left loop, starting below the voltage source:
−vg + vi = 0 so 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.
2–2 CHAPTER 2. Circuit Elements
AP 2.
[a] Note from the circuit that vx = − 25 V. To find α write a KCL equation at the top left node, summing the currents leaving: 15 A + αvx = 0 Substituting for vx,
15 A + α(− 25 V) = 0 so α(25 V) = 15 A
Thus α =
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. To find this current, write a KCL equation at the top left node, summing the currents leaving the node:
−αvx + iv = 0 so 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] A KVL equation gives
−vg + vR = 0 so vR = vg = 1 kV Note from the circuit that the current through the resistor is ig = 5 mA. Use Ohm’s law to calculate the value of the resistor:
R =
vR ig
1 kV 5 mA
= 200 kΩ
Using the passive sign convention to calculate the power in the resistor,
pR = (vR)(ig) = (1 kV)(5 mA) = 5 W The resistor is dissipating 5 W of power.
2–4 CHAPTER 2. Circuit Elements
[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
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
Problems 2–
[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
Also use Ohm’s law to calculate the value of the current through the 24 Ω resistor:
i 2 =
120 V
= 5 A
Now write a KCL equation at the top middle node, summing the currents leaving:
−i 1 + i 2 + i 3 = 0 so i 1 = i 2 + i 3 = 5 + 15 = 20 A
Write a KVL equation clockwise around the left loop, starting below the voltage source:
− 200 V + v 1 + 120 V = 0 so v 1 = 200 − 120 = 80 V
Problems 2–
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 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^2 t 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 30 i 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.
2–8 CHAPTER 2. Circuit Elements
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 + 189, 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
Thus,
v = 240, 000 i 1 − 8 V = 240,000(25 × 10 −^6 ) − 8 V = 6 V − 8 V = − 2 V
We now know the values of voltage and current for every circuit element. Let’s construct a power table:
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
Problems 2–
Problems
P 2.
Vbb = no-load voltage of battery
Rbb = internal resistance of battery Rx = resistance of wire between battery and switch
Ry = resistance of wire between switch and lamp A
Ra = resistance of lamp A Rb = resistance of lamp B
Rw = resistance of wire between lamp A and lamp B
Rg 1 = resistance of frame between battery and lamp A Rg 2 = resistance of frame between lamp A and lamp B
S = switch
P 2.2 Since we know the device is a resistor, we can use Ohm’s law to calculate the resistance. From Fig. P2.2(a),
v = Ri so R =
v i Using the values in the table of Fig. P2.2(b),
R =
= 8kΩ
P 2.3 The resistor value is the ratio of the power to the square of the current:
500 12
P 2.4 Since we know the device is a resistor, we can use the power equation. From Fig. P2.4(a),
p = vi =
v^2 R
so R =
v^2 p
Using the values in the table of Fig. P2.4(b)
R =
(−8)^2
(−4)^2
(4)^2
(8)^2
(12)^2
(16)^2
Problems 2–
KVL, outer loop: 60 V − 100 V − v 5 − v 25 = 0
Note that since v 5 , v 20 , and v 25 are not specified, we can choose values that satisfy the equations. For example, let v 5 = − 80 V, v 20 = 40V, and v 25 = 40V. There are many other voltage values that will satisfy the equations, too. Thus, the interconnection is valid because it does not violate Kirchhoff’s laws. We can now calculate the power developed by the two voltage sources:
pv−sources = p 60 + p 100 = −(60)(5) + (100)(5) = 200 W.
Since the power is positive, the sources are absorbing 200 W of power, or developing − 200 W of power.
P 2.
Write the two KCL equations, summing the currents leaving the node:
KCL, top node: − 30 A − i 8 + 10A = 0A
KCL, bottom node: 30 A + i 8 − 10 A = 0A
Note that the value i 8 = − 20 A satisfies these two equations. Write the three KVL equations, summing the voltages in a clockwise direction:
KVL, left loop: − v 30 − 16 V + 8V = 0
KVL, right loop: − 10 V + v 10 − 8 V = 0
KVL, outer loop: − 16 V − 10 V + v 10 − v 30 = 0
Note that v 30 = − 8 V and v 10 = 18V satisfy the three KVL equations. The interconnection is valid, since neither of Kirchhoff’s laws is violated. We use the values of i 8 , v 30 and v 10 stated above to calculate the power associated with each source:
p30A = −(30)(−8) = 240 W p16V = −(30)(16) = − 480 W
2–14 CHAPTER 2. Circuit Elements
p8V = −(−20)(8) = 160 W p10V = −(10)(10) = − 100 W
p10A = (10)(18) = 180 W ∑ Pabs =
∑ Pdel = 580 W
Power developed by the current sources:
pi−sources = p30A + p10A = 240 + 180 = 420 W
Since power is positive, the sources are absorbing 420 W of power, or developing − 420 W of power.
P 2.8 The interconnect is valid since it does not violate Kirchhoff’s laws.
−10 + 40 + v5A − 50 = 0 so v5A = 20 V (KVL)
15 + 5 + i50V = 0 so i50V = − 20 A (KCL)
p15A = −(15)(50) = − 750 W p50V = (20)(50) = 1000 W
p5A = −(5)(20) = − 100 W p10V = (5)(10) = 50 W
p40V = −(5)(40) = − 200 W ∑ Pdev =
∑ Pabs = 1050 W
P 2.9 First there is no violation of Kirchhoff’s laws, hence the interconnection is valid. Kirchhoff’s voltage law requires
−20 + 60 + v 1 − v 2 = 0 so v 1 − v 2 = − 40 V
The conservation of energy law requires
−(5 × 10 −^3 )v 2 − (15 × 10 −^3 )v 2 − (20 × 10 −^3 )(20) + (20 × 10 −^3 )(60) + (20 × 10 −^3 )v 1 = 0
or
v 1 − v 2 = − 40 V
Hence any combination of v 1 and v 2 such that v 1 − v 2 = − 40 V is a valid solution.
2–16 CHAPTER 2. Circuit Elements
Substitute the value v∆ = 10 V into the second KVL equation and find vdep = 30 V. Substitute the value v∆ = 10 V into the third equation and find v 75 = − 40 V. These values satisfy the first equation. Thus, the interconnection is valid because it does not violate Kirchhoff’s laws. Use the values for v∆, v 75 , and vdep above to calculate the total power developed in the circuit:
p50V = (75)(50) = 3750 W p75A = (75)(−40) = − 3000 W
p20V = 5(10) = 1000 W pds = −(50)(30) = − 1500 W
p25A = −(25)(10) = − 250 W ∑ Pdev = 3750 + 1000 = 4750 W =
∑ Pabs
P 2.12 [a] Yes, Kirchhoff’s laws are not violated. (Note that i∆ = − 8 A.)
[b] No, because the voltages across the independent and dependent current sources are indeterminate. For example, define v 1 , v 2 , and v 3 as shown:
Kirchhoff’s voltage law requires v 1 + 20 = v 3
v 2 + 100 = v 3
Conservation of energy requires −8(20) − 8 v 1 − 16 v 2 − 16(100) + 24v 3 = 0 or v 1 + 2v 2 − 3 v 3 = − 220 Now arbitrarily select a value of v 3 and show the conservation of energy will be satisfied. Examples: If v 3 = 200 V then v 1 = 180 V and v 2 = 100 V. Then
180 + 200 − 600 = − 220 (CHECKS)
If v 3 = − 100 V, then v 1 = − 120 V and v 2 = − 200 V. Then
− 120 − 400 + 300 = − 220 (CHECKS)
Problems 2–
P 2.13 First, 10 va = 5 V, so va = 0. 5 V KVL for the outer loop: 5 − 20 + v9A = 0 so v9A = 15 V KVL for the right loop: 5 − 0 .5 + vg = 0 so vg = − 4. 5 V KCL at the top node: 9 + 6 + ids = 9 so ids = − 15 A Thus,
p9A = −(9)(15) = − 135 W p20V = (9)(20) = 180 W
pvg = −(6)(− 4 .5) = 27 W p6A = (6)(0.5) = 3 W
pds = −(15)(5) = − 75 W
∑ Pdev =
∑ Pabs = 210 W
P 2.
[a] Write a KVL equation clockwise aroud the right loop, starting below the 300 Ω resistor:
−va + vb = − 0 so va = vb
Using Ohm’s law,
va = 300ia and vb = 75ib Substituting,
300 ia = 75ib so ib = 4ia
Write a KCL equation at the top middle node, summing the currents leaving:
−ig + ia + ib = 0 so ig = ia + ib = ia + 4ia = 5ia
Write a KVL equation clockwise around the left loop, starting below the voltage source: − 200 V + v 40 + va = 0
From Ohm’s law,
v 40 = 40ig and va = 300ia
Problems 2–
P 2.
[a] Write a KVL equation clockwise around the right loop:
−v 60 + v 30 + v 90 = 0 From Ohm’s law,
v 60 = (60 Ω)(4 A) = 240 V, v 30 = 30io, v 90 = 90io
Substituting,
− 240 V + 30io + 90io = 0 so 120 io = 240 V
Thus io =
240 V
= 2 A
Now write a KCL equatiohn at the top middle node, summing the currents leaving:
−ig + 4 A + io = 0 so ig = 4 A + 2 A = 6 A
[b] Write a KVL equation clockwise around the left loop:
−vo + v 60 = 0 so vo = v 60 = 240 V
[c] Calculate power using p = vi for the source and p = Ri^2 for the resistors: psource = −voig = −(240 V)(6 A) = − 1440 W
p60Ω = 4^2 (60) = 960 W
p30Ω = 30i^2 o = (30)2^2 = 120 W
p90Ω = 90i^2 o = (90)2^2 = 360 W ∑ Pdev = 1440 W
∑ Pabs = 960 + 120 + 360 = 1440 W
2–20 CHAPTER 2. Circuit Elements
P 2.17 [a]
v 2 = 2(20) = 40 V v8Ω = 80 − 40 = 40 V i 2 = 40 V/8 Ω = 5 A i 3 = io − i 2 = 2 − 5 = − 3 A v4Ω = (−3)(4) = − 12 V v 1 = 4i 3 + v 2 = −12 + 40 = 28 V i 1 = 28 V/4 Ω = 7 A [b] i 4 = i 1 + i 3 = 7 − 3 = 4 A p13Ω = 42 (13) = 208 W
p8Ω = (5)^2 (8) = 200 W
p4Ω = 72 (4) = 196 W p4Ω = (−3)^2 (4) = 36 W
p20Ω = 22 (20) = 80 W
[c]
∑ Pdis = 208 + 200 + 196 + 36 + 80 = 720 W
ig = i 4 + i 2 = 4 + 5 = 9 A
Pdev = (9)(80) = 720 W
