Circuits with Dependent
CIRCUITS WITH DEPENDENT
− I1 3 kΩ
12 V 5 kΩ
VA = 2000I1
Vo +−
Applying KVL, we obtain
−12 + 3kI1 − VA + 5kI1 = 0 where
VA = 2000I1
and the units of the multiplier, 2000, are ohms. Solving these equations yields I1 = 2 mA
Then
Vo = (5 k)I1
= 10 V SOLUTION
Next let us consider the case in which an independent voltage source is connected between two nonreference nodes.
Given the circuit in Fig. 2.39 containing a current-controlled current source, let us find the voltage Vo.
EXAMPLE 2.28
Figure 2.39 Circuit used in Example 2.28.
Vo
VS 4 Io Io
4 kΩ
3 kΩ 2 kΩ
10 mA
+ –
+
–
CIRCUITS WITH DEPENDENT SOURCES
Applying KCL at the top node, we obtain
10 × 10−3 + _______ VS
2k + 4k + ___ VS
3k − 4mIo = 0 where
Io = ___ VS
3k
Substituting this expression for the controlled source into the KCL equation yields 10−2 + ___ VS
6k + ___ VS
3k − ___ 4VS
3k = 0 Solving this equation for VS, we obtain
VS = 12 V
The voltage Vo can now be obtained using a simple voltage divider; that is, Vo =
[
4k — 2k + 4k]
VS= 8 V
SOLUTION
The network in Fig. 2.40 contains a voltage-controlled voltage source. We wish to find Vo
in this circuit.
EXAMPLE 2.29
Figure 2.40 Circuit used in Example 2.29.
+−
+−
Vo +
– I 3 kΩ
12 V 1 kΩ
2Vo
Applying KVL to this network yields
−12 + 3kI + 2Vo + 1kI = 0 where
Vo = 1kI Hence, the KVL equation can be written as
−12 + 3kI + 2kI + 1kI = 0 or
I = 2 mA Therefore,
Vo = 1kI
= 2 V
SOLUTION
An equivalent circuit for a FET common-source amplifier or BJT common-emitter amplifier can be modeled by the circuit shown in Fig. 2.41a. We wish to determine an expression for the gain of the amplifier, which is the ratio of the output voltage to the input voltage.
EXAMPLE 2.30
R1
R3 R4 R5
(a) i1(t)
υi(t) gm υg(t) υo(t)
+
− R2
υg(t) +
–
R1
RL
(b) i1(t)
υi(t) gmυg(t) υo(t)
+
– R2
υg(t) +
–
+−
+− Figure 2.41
Example circuit containing a voltage-controlled current source.
Note that although this circuit, which contains a voltage-controlled current source, appears to be somewhat complicated, we are actually in a position now to solve it with techniques we have studied up to this point. The loop on the left, or input to the amplifier, is essentially detached from the output portion of the amplifier on the right. The voltage across R2 is υg(t), which controls the dependent current source.
To simplify the analysis, let us replace the resistors R3, R4, and R5 with RL such that 1
— RL = 1 —
R3 + 1 —
R4 + 1 —
R5
Then the circuit reduces to that shown in Fig. 2.41b. Applying Kirchhoff’s voltage law to the input portion of the amplifier yields
υi(t) = i1(t)(R1 + R2) and
υg(t) = i1(t)R2
Solving these equations for υg(t) yields
υg(t) = _______ R1 + RR2 2 υi(t)
From the output circuit, note that the voltage υo(t) is given by the expression υo(t) = −gmυg(t)RL
Combining this equation with the preceding one yields υo(t) = −g________ R1m + RRLR22 υi(t)
Therefore, the amplifier gain, which is the ratio of the output voltage to the input voltage, is given by
υo(t)
____ υi(t) = − _______ Rgm1 + RRLR22 SOLUTION
At this point it is perhaps helpful to point out again that when analyzing circuits with depen-dent sources, we first treat the dependepen-dent source as though it were an independepen-dent source when we write a Kirchhoff’s current or voltage law equation. Once the equation is written, we then write the controlling equation that specifies the relationship of the dependent source to the unknown variable. For instance, the first equation in Example 2.28 treats the dependent source like an independent source. The second equation in the example specifies the relationship of the dependent source to the voltage, which is the unknown in the first equation.
Reasonable values for the circuit parameters in Fig. 2.41a are R1 = 100 Ω, R2= 1 kΩ, gm = 0.04 S, R3 = 50 kΩ, and R4 = R5 = 10 kΩ. Hence, the gain of the amplifier under these conditions is
υ____ o(t)
υi(t) = −(0.04)(4.545)(10——— (1.1)(103)3)(1)(103)
= −165.29 Thus, the magnitude of the gain is 165.29.
E2.27 Find Vo in the circuit in Fig. E2.27.
+−
Vo VA
I 2VA
4 kΩ 6 V 8 kΩ
+
−
+ −
+−
Figure E2.27
E2.28 Find Vo in the network in Fig. E2.28.
Vo VS
Vo 2000 2 mA
2 kΩ 1 kΩ
6 kΩ +
– +
–
—
Figure E2.28
E2.29 Find VA in Fig. E2.29.
10kΩ 5kΩ
36 V 12 V
5kΩ VA
Vx 2Vx
+
+
−
−
+– +
– + –
Figure E2.29
ANSWER:
Vo = 12 V.
ANSWER:
Vo = 8 V.
ANSWER:
VA = −12 V.
LEARNING ASSESSMENTS
E2.30 Find V1 in Fig. E2.30.
4kΩ
8kΩ
50 V 18 V
8kΩ V1
Vx
0.5 Vx
+
+
–
–
+– +–
+−
Figure E2.30
ANSWER:
V1 = −32/3 V.
E2.31 Find Ix in Fig. E2.31.
10 mA 2Ix 2kΩ 3 mA 5kΩ 10kΩ
Ix
Figure E2.31
E2.32 Find Vo in Fig. E2.32.
6 mA 6kΩ
4kΩ 0.5Ix
Ix
12kΩ Vo +
Figure E2.32 −
E2.33 If the power supplied by the 3-A current source in Fig. E2.33 is 12 W, find VS and the power supplied by the 10-V source.
3 A
6Ω 10 V
4Ω
3Ω 4Ω
VS + 5Ω
– +
– Figure E2.33
ANSWER:
Ix = −1.5 mA.
ANSWER:
Vo = 16 V.
ANSWER:
VS = 42 V;
−30 W.
S U M M A R Y
■ Ohm’s law V = IR
■ The passive sign convention with Ohm’s law The current enters the resistor terminal with the positive voltage refer-ence.
■ Kirchhoff’s current law (KCL) The algebraic sum of the cur-rents leaving (entering) a node is zero.
■ Kirchhoff’s voltage law (KVL) The algebraic sum of the voltages around any closed path is zero.
■ Solving a single-loop circuit Determine the loop current by applying KVL and Ohm’s law.
■ Solving a single-node-pair circuit Determine the voltage between the pair of nodes by applying KCL and Ohm’s law.
■ The voltage-division rule The voltage is divided between two series resistors in direct proportion to their resistance.
■ The current-division rule The current is divided between two parallel resistors in reverse proportion to their resistance.
■ The equivalent resistance of a network of resistors Combine resistors in series by adding their resistances.
Combine resistors in parallel by adding their conductances.
The wye-to-delta and delta-to-wye transformations are also an aid in reducing the complexity of a network.
■ Short circuit Zero resistance, zero voltage; the current in the short is determined by the rest of the circuit.
■ Open circuit Zero conductance, zero current; the voltage across the open terminals is determined by the rest of the circuit.
P R O B L E M S
2.1 Determine the current and power dissipated in the resistor in Fig. P2.1.
9 V +− 12 Ω
Figure P2.1
2 mA Rx
Figure P2.2
Figure P2.3
2.4 In the network in Fig. P2.4, the power absorbed by Gx is 20 mW. Find Gx.
Figure P2.4
2.2 Determine the voltage across the resistor in Fig. P2.2 and the power dissipated.
2.3 In the network in Fig. P2.3, the power absorbed by Rx is 20 mW. Find Rx.
2.5 A model for a standard two D-cell flashlight is shown in Fig. P2.5. Find the power dissipated in the lamp.
1-Ω lamp
1.5 V
1.5 V
2.6 An automobile uses two halogen headlights connected as shown in Fig. P2.6. Determine the power supplied by the battery if each headlight draws 3 A of current.
12 V
+ −
Figure P2.6
2.7 Many years ago a string of Christmas tree lights was manu-factured in the form shown in Fig. P2.7a. Today the lights are manufactured as shown in Fig. P2.7b. Is there a good reason for this change?
(a)
(b) Figure P2.7
2 A 12 Ω
12 mA Gx
2.8 Find I1, I2, and I3 in the network in Fig. P2.8.
5 A
3 A A
6 A
2 A 4 A 8 A
I2 4 A
I3 I1
B
C
Figure P2.9
2.10 Find I1 in the network in Fig. P2.10.
Figure P2.10
2.9 Find I1 in the network in Fig. P2.9.
Figure P2.11
2.12 Find Io and I1 in the circuit in Fig. P2.12.
Figure P2.12
2.11 Find I1 in the circuit in Fig. P2.11.
20 mA
6 mA
I1 4 mA
2 mA 6 mA
I1
+−
4 mA
2 mA
12 mA
I1
+−
5 mA
2 mA
4 mA
Io 3 mA I1
− + +−
Figure P2.13
2.14 Find Ix in the circuit in Fig. P2.14.
Figure P2.14
2.13 Find Ix, Iy,and Iz in the network in Fig. P2.13.
2.15 Find Ix in the network in Fig. P2.15.
Figure P2.15
2.16 Find I1 in the network in Fig. P2.16.
Figure P2.16
2.17 Find Vbd in the circuit in Fig. P2.17.
4 mA 3 mA
2 mA 12 mA
Ix
Iy
Iz
− +
2 mA 10 mA
Ix
5Ix
12 mA 4 mA
2Ix 3Ix
Ix
2 mA 4 mA
2Ix
Ix I1
Figure P2.17
a b c
d
− +
12 V
4 V 6 V
2 V +
+
−
−
+−
2.35 Find the power absorbed by the dependent source in the circuit in Fig. P2.35.
Figure P2.35
2.42 Calculate the power absorbed by the dependent source in the circuit in Fig. P2.42.
Figure P2.42
2.43 Find VA and Vo in the circuit in Fig. P2.43.
Figure P2.43
2.44 Find Vo and the power absorbed by the 2 kΩ resistor in Fig. P2.44.
Figure P2.44
2.45 Find the power absorbed or supplied by the 12-V source in the network in Fig. P2.45.
2.46 Find Vo in the circuit in Fig. P2.46.
2.47 Find Io in the network in Fig. P2.47.
Figure P2.47
2.48 Find Io in the network in Fig. P2.48.
Figure P2.48
2.49 Find the power supplied by each source in the circuit in Fig. P2.49.
2.52 Find Io in the circuit in Fig. P2.52.
2.60 Find RAB in the circuit in Fig. P2.60.
2.68 Given the resistor configuration shown in Fig. P2.68, find the equivalent resistance between the following sets of ter-minals: (1) a and b, (2) b and c, (3) a and c, (4) d and e, (5) a and e, (6) c and d, (7) a and d, (8) c and e, (9) b and d, and (10) b and e.
Figure P2.68
4 Ω 4 Ω
10 Ω
12 Ω a
5 Ω 5 Ω
b
c
d e
Figure P2.69
2.69 Determine the total resistance, RT, in the circuit in Fig. P2.69.
2.70 Determine the total resistance, RT, in the circuit in Fig. P2.70.
Figure P2.70 2 kΩ
1 kΩ 1 kΩ
12 kΩ
12 kΩ
12 kΩ 12 kΩ 12 kΩ
12 kΩ 12 kΩ
12 kΩ
12 kΩ
12 kΩ 12 kΩ 4 kΩ
12 kΩ
12 kΩ RT
3 kΩ
3 kΩ
12 kΩ
12 kΩ
12 kΩ
12 kΩ
12 kΩ 6 kΩ 6 kΩ
6 kΩ
16 kΩ
12 kΩ 12 kΩ 2 kΩ
16 kΩ
RT
2.71 Determine the total resistance, RT, in the circuit in Fig. P2.71.
Figure P2.71 24 kΩ
8 kΩ
24 kΩ
8 kΩ
18 kΩ 9 kΩ
9 kΩ
1 kΩ 24 kΩ
RT 6 kΩ
2.73 Find I1 and Vo in the circuit in Fig. P2.73. 2.72 Find the power supplied by the source in the network in
Fig. P2.72. All resistors are 12 kΩ.
Figure P2.72
12 A
2.80 Find Vab in the network in Fig. P2.80. circuit in Fig. P2.86.
Figure P2.86
2.90 In the network in Fig. P2.90, V1 = −14 V. Find VS.
2.87 Find the power supplied by the current source in the net-work in Fig. P2.87. All resistors are 12 Ω.
Figure P2.87
2.95 Find the value of VS in the network in Fig. P2.95 such that the power supplied by the current source is 0.
Figure P2.95 network in Fig. P2.99.
Figure P2.99
2.100 The 40-V source in the circuit in Fig. P2.100 is absorbing 80 W of power. Find Vx.
2.106 Find the value of Vx in the circuit in Fig. P2.106 such that the power supplied by the 5-A source is 60 W.
Figure P2.106
2.107 Find the power absorbed by the network in Fig. P2.107.
Figure P2.107 that the power supplied by the 3-A source is 20 W.
gIx that the 5-A current source supplies 50 W.
+−
2.109 Find the power supplied by the 24-V source in the circuit
2.115 Find the power supplied by the 6-mA source in the net-work in Fig. P2.115.
Figure P2.115
2.121 Find Vo in the circuit in Fig. P2.121.
Figure P2.121
2.122 A typical transistor amplifier is shown in Fig. P2.122.
Find the amplifier gain G (i.e., the ratio of the output volt-age to the input voltvolt-age).
Figure P2.122
2Iy Vo
6 A
Vx Iy 2Ω
1Ω
− +
2 Ω Vx
2
+–
VS = 250 mV
100 Ω
300 Ω 4 kΩ
4 × 105 Ib
5 kΩ 500 Ω Vo
+
Ib −
+–
2.117 Find Vo in the network in Fig. P2.117.
Figure P2.117
2.118 Find I1 in the network in Fig. P2.118.
Figure P2.118
2.119 A single-stage transistor amplifier is modeled as shown in Fig. P2.119. Find the current in the load RL.
Figure P2.119 I +−
24 V
2 kΩ
4 kΩ 2 Vo
Vo
+
−
+−
60 V
8 Ω
6 Ω
4 Ω 12 Ω 4.5 A
I1 I2
3 I2
+−
VS = 250 mV 100 Ib
RS = 1 kΩ
Rb = 250 Ω
Ro = 4 kΩ RL = 400 Ω
+
− Ib Io
2.120 Find Io in the circuit in Fig. P2.120.
Figure P2.120
2Vx
Vx Io
Vo
2 A 2Ω
1Ω 4 A
− +
2.123 Find Vx in the network in Fig. P2.123.
Ix
Vx 6 Ω
3 Ω 6 Ω
3 Ω
2 Ω 1 Ω
4 A +
−
2Ix Vo
Figure P2.123
Figure P2.124
2.124 Find Vo in the network in Fig. P2.124.
2Vy 6 A
Vo
Vy
4Vx 2Ω
2Ω
− + Vx
1Ω
2Ω
−
+
2.125 Find I1, I2, and I3 in the circuit in Fig. P2.125.
Figure P2.125
2.126 Find Io in the network in Fig. P2.126.
Figure P2.126
2.127 Find the power absorbed by the 12-kΩ resistor on the right side of the network in Fig. P2.127.
Figure P2.127
+− 8 Ω 2 A 24 V
4I1
I2
I1 I3
+−
Io
8 Ω
4 Ω Vx
6 Ω 5 A 3 Vx
+
−
5 mA
4 kΩ
2 kΩ
12 kΩ
12 kΩ
Vx 3 kΩ
Vo +
+
− −
Vx 2000
2.128 Find the power absorbed by the 12-kΩ resistor in the net-work in Fig. P2.128.
6 mA
6 kΩ 3 kΩ
4 kΩ 12 kΩ
4 kΩ
Vo +
− 3Io
Io
2FE-1 What is the power generated by the source in the network in Fig. 2PFE-1?
a. 2.8 W c. 3.6 W b. 1.2 W d. 2.4 W
6 kΩ
120 V
5 kΩ
18 kΩ
4 kΩ 6 kΩ
12 kΩ
+−
Figure 2PFE-1
T Y P I C A L P R O B L E M S F O U N D O N T H E F E E X A M
2FE-2 Find Vab in the circuit in Fig. 2PFE-2.
a. −5 V c. 15 V b. 10 V d. −10 V
5 Ω
15 Ω 10 Ω
10 Ω
a
b 4 A
Vab
Figure 2PFE-2 2.129 Find the value of k in the network in Fig. P2.129 such
that the power supplied by the 6-A source is 108 W.
Figure P2.129 6 A
6 Ω 3 Ω
6 Ω 12 Ω
4 Ω
kIo
Io
2.130 If the power absorbed by the 10-V source in Fig. P2.130 is 40 W, calculate IS.
Figure P2.130
0.6Vx +− + 10 V
Vx −
Is 6 Ω
10 Ω 15 Ω
4 Ω 5 Ω
+
−
2.131 If the power supplied by the 2-A current source in Fig. P2.131 is 50 W, calculate k.
Figure P2.131 50 V + 2 A
− kI1
2 Ω
5 Ω 2 Ω
5 Ω I1 4 Ω
2FE-3 If Req = 10.8 Ω in the circuit in Fig. 2PFE-3, what is R2?
2FE-4 Find the equivalent resistance of the circuit in Fig. 2PFE-4 at the terminals A-B.
a. 4 kΩ c. 8 kΩ net-work in Fig. 2PFE-5. What is R?
a. 17.27 Ω c. 19.25 Ω
2FE-6 Find the power supplied by the 40-V source in the circuit in Fig. 2PFE-6.
2FE-10 Find the current Ix in Fig. 2PFE-10.
a. 1/2 A c. 3/2 A b. 5/3 A d. 8/3 A
1 Ω 1 Ω
10 Ω 8 Ω 3 Ω
2 Ω 12 V
Ix
+−
Figure 2PFE-10 2FE-9 What is the voltage Vo in the circuit in Fig. 2PFE-9?
a. 2 V c. 5 V b. 8 V d. 12 V
2 Ω 3 Ω
1 Ω 4 A
2 A Vo
+
− Figure 2PFE-9