Basic Engineering Circuit Analysis
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Transcript of Basic Engineering Circuit Analysis
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Basic Engineering Circuit Analysis
CH#2
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A variable resistor is a potentiometer with only two connecting wires instead of three. However, although the actual component is the same, it does a very different job. The pot allows us to control the potential passed through a circuit. The variable resistance lets us adjust the resistance between two points in a circuit.
A variable resistance is useful when we don't know in advance what resistor value will be required in a circuit. By using pots as an adjustable resistor we can set the right value once the circuit is working. Controls like this are often called 'presets' because they are set by the manufacturer before the circuit is sent to the customer. They're usually hidden away inside the case of the equipment, away from the fingers of the users!
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Q#2.37: Given the network in Fig. P2.37, we wish to obtain a voltage of 2 V ≤ V0 ≤ 9 V across the full range of the potentiometer. Determine the values of R1 and R2.Solution:Circuit diagram:
R1
10 V 1 kΩ+
V0
R2
-
P 2.37
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case A case B
Wiper at bottom of variable R
According to voltage divider rule
R2
V0(min) = × 10 V = 2 V R1 + R2 + 1000Equation (i)
Wiper at top of variable R
According to voltage divider rule
R2 + 1000V0(max) = × 10 V = 9 V R1 + R2 + 1000Equation (ii)
V0(max) R2 + 1000 =
V0(min) R2
9 R2 + 1000 = 2 R2
2R2 + 2000 = 9R2
7R2 = 2000
R2 = 285.714 Ω
Substituting the value of R2 in Equation (i)
285.7142 V = × 10 V R1 + 285.714 + 1000
2[R1 + 285.714 + 1000] = 10[285.714]2R1 + 571.428 + 2000 = 2857.142R1 = 2857.14 - 571.428 – 2000
R1 = 142.856 Ω
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Q#2.41: Find V0 in the network in Fig. P2.41.Solution:Circuit diagram:
A
1 Ω
+
12 A9 Ω 2 Ω V0
- B
18 Ω
Fig. (a)
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A
1 Ω
12 A 18 Ω 9 Ω 2 Ω
B Fig. (b)Series combination= 1 Ω + 2 Ω= 3 Ω
Parallel combination
9 Ω × 18 Ω= 9 Ω + 18 Ω
162 Ω × Ω= 27 Ω
= 6 Ω A
6 Ω 3 Ω 12 A
B Fig. (c)
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Parallel combination
3 Ω × 6 Ω= 3 Ω + 6 Ω
18 Ω × Ω= 9 Ω
= 2 Ω
A
2 Ω 12 A
B Fig. (d)According to ohm’s LawVAB = (12 A)(2 Ω )
VAB = 24 Volts
From Fig. (b)According to Voltage divider rule:
2 ΩV0 = × VAB
1 Ω + 2 Ω
2 ΩV0 = × 24 V 3 Ω
V0 = -16 Volts based on polarity
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A
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Q#2.69: Find the value of Vx in the network in Fig. P2.69 such that the 5-A current source supplies 50 watt.Solution:Circuit diagram:
- +I3
2 Ω Vx
I5 = 5 A + - - +
I2 4 Ω + 4 Ω +
5 V 2 Ω 2 Ω 5 A
- -
I4
Fig. (a)Pcurrent-source = V current-source(5 A)HerePcurrent-source = 50 watts
50 wattsVcurrent-source =
5 A
Vcurrent-source = 10 Volts
According to ohm’s LawV4Ω = I4Ω (4Ω )HereI4Ω = 5 ATherefore,V4Ω = (5 A)(4Ω )
V4Ω = 20 Volts
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4 Ω - +
+ +
2 Ω 5 A
- -
Fig. (b)According to KVLSum of all the voltage drop = sum of all the voltage riseVcurrent-source = V4Ω + V2Ω
V2Ω = Vcurrent-source - V4Ω
Substituting the corresponding valuesV2Ω = 10 Volts – 20 Volts
V2Ω = -10 Volts
Following Fig. (A)According to ohm’s Law V2Ω
I4 = 2 Ω
-10 VI4 = 2 Ω
I4Ω = -5 A
V4Ω = 5 V - VA
HereVA = V2Ω = -10 VoltsV4Ω = 5 V – (-10 V)
V4Ω = 15 V
V4Ω
I2 = 4 Ω
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15 VI2 = 4 Ω
I2 = 3.75 AApplying KCL at node AI5 + I2 = I4 + I3
Substituting the corresponding values5 A + 3.75 A = -5 A + I3
I3 = 13.75 A
- +I3
2 Ω Vx
I5 = 5 A + - - +
I2 4 Ω + 4 Ω +
5 V 2 Ω 2 Ω 5 A
- -
I4
Fig. (c)
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- +I3
2 Ω Vx
+ - Fig. (d)
I2
4 ΩAccording to KVLSum of all the voltage drop = sum of all the voltage riseVx = I3(2 Ω ) + I2(4 Ω )Vx = (13.75 A)(2 Ω ) + (3.75 A)(4 Ω )Vx = 27.5 V + 15 V
Vx = 42.5 V
Q#2.70: Find the value of V1 in the network in Fig. P2.70 such that Va = 0.Solution:Circuit diagram:
8 V 2 Ω I1 A I2 B I4
+ Ia I3
2 Ω Va 2 Ω 4 Ω 2 Ω V1
V1 -
According to ohm’s LawVa = Ia(2 Ω )If Va = 0, then Ia = 0Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI1 = Ia + I2 … (i)Here
V1 Va
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I1
According to ohm’s Law
V1 - Va
I1 = 2 Ω
HereVa = 0
V1 – 0 I1 =
2 Ω
V1
I1 = 2 Ω
from equation (i)I1 = o A + I2
I1 = I2
Substituting the corresponding value of I1
V1
I2 = … (ii) 2 ΩApplying KCL at Node Labeled ‘B’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI2 = I3 + I4 … (iii)VB – VA = 8 VHereVA = 0 VVB – 0 = 8 V
VB = 8 V
According to ohm’s Law VB
I3 = 4 Ω
8 VI3 =
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4 Ω
I3 = 2 A
VB V1
I4
According to ohm’s Law
VB – V1
I1 = 2 Ω
HereVB = 8 V
8 V – V1
I4 = 2 Ω
Substituting the corresponding values of I3 & I4 in equation (iii)
8 V – V1
I2 = 2 A + … (iv) 2 ΩComparing equations (ii) & (iv)V1 8 V – V1
= 2 A + 2 Ω 2 Ω
V1 8 V – V1
- = 2 A2 Ω 2 Ω
V1 – [8 V – V1]= 2 A
2 Ω
V1 – 8 V + V1 = 4 V2V1 = 12 V
V1 = 6 V
Q#2.71: Find the value of Vx in the circuit in Fig. P2.71 such that the power supplied by the 5-A source is 60 W.Solution:
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Circuit diagram:1 Ω
- +
Vx -
1 Ω 3 A
+I2
A B
+ 4 Ω I3 +
5 V 2 Ω 2 Ω 5 A
- -
I1
Fig. (a)Pcurrent-source = V current-source(5 A)HerePcurrent-source = 60 watts
60 wattsVcurrent-source =
5 A
Vcurrent-source = 12 Volts = VB
According to ohm’s Law VB
I1 = 2 Ω
12 VI1 = 2 Ω
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I1 = 6 A
VA VB
I3
According to ohm’s Law
VB – VA
I3 = 4 Ω
HereVA = 5 V, VB = 12 V
12 – 5I3 =
4 Ω
7 VI3 = 4 Ω
I3 = 1.75 A
Applying KCL at Node Labeled ‘B’ Sum of all the currents entering into the junction = sum of all the currents leaving that junction5 A + 3 A = I1 + I2 + I3
Substituting the corresponding values of I1 & I3
5 A + 3 A = 6 A + I2 + 1.75 A8 A = 7.75 A + I2
I2 = 0.25 A
Applying KVL around the dotted pathSum of all the voltage rise = sum of all the voltage dropI2(1) + I2(1) = Vx + I3(4)2I2 = Vx + 4I3
Substituting the corresponding values of I2 & I3
2(0.25) = Vx + 4(1.75)0.5 = Vx + 7
Vx = -6.5 V
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Q#2.22: Find V0 in the circuit in Fig. P2.22.Solution:Circuit diagram: Vx
- +
+ 4 Ω - + 12 Ω - 12 V actual polarity
2Ix + V0 - Vx
Ix
Fig. (a)According to KVLSum of all the voltage rise = sum of all the voltage drop2Ix + 12 = 4(Ix) + 12(Ix) + Vx
HereVx = -12(Ix)2Ix + 12 = 4(Ix) + 12(Ix) + (-12(Ix))-2Ix = -12
Ix = 6 A
Applying KVL around the dotted pathSum of all the voltage rise = sum of all the voltage drop12 + V0 = 12Ix
Substituting the corresponding value of Ix
12 + V0 = 12(6)12 + V0 = 72
V0 = 60 VoltsQ#2.12: Find I0 and I1 in the circuit in Fig. P2.12. Solution:Circuit diagram:
5 mA
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4 mA
A I1 B
2 mA+
-
I0 3 mA
C I2
Applying KCL at Node Labeled ‘B’Sum of all the currents entering into the junction = sum of all the currents leaving that junction4 mA + 2 mA = I2
I2 = 6 mA
Applying KCL at Node Labeled ‘C’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI2 = I0 + 3 mAHere I2 = 6 mA 6 mA = I0 + 3 mAI0 = 3 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI0 = I1 + 5 mAHere I0 = 3 mA 3 mA = I1 + 5 mAI1 = -2 mA
Q#2.97: Given that V0 = 4 V in the network in Fig. P2.97, find VS.Solution:Circuit diagram:
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C - + + - +
I1 AVS 6 V 3 kΩ 2 kΩ
I2 I0
- + 2 mA V0 = 4 V 3 kΩ 12 kΩ 1 kΩ
+ -
I3
- B
Fig. (a)
+ - V0 VB
I0
According to ohm’s Law
V0 – VB
I0 = 1 kΩ
HereV0 = 4 V, VB = 0 V
4 – 0I0 =
1 kΩ
4 VI0 = 1 kΩ
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I0 = 4 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junction2 mA = I1 + I0
HereI0 = 4 mA2 mA = I1 + 4 mA
I1 = -2 mA
Applying KVL around the dotted pathSum of all the voltage rise = sum of all the voltage drop12000I2 + 3000I1 = 6 + 3000I0
HereI1 = -2 mAI0 = 4 mA12000I2 + 3000(-2 mA) = 6 + 3000(4 mA)12000I2 - 6 = 6 + 1212000I2 = 24
I2 = 2 mA
Applying KCL at Node Labeled ‘C’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI1 + I3 = I2
HereI1 = -2 mAI2 = 2 mA-2 mA + I3 = 2 mA
I3 = 4 mA
C - + + - +
I1 AVS 6 V 3 kΩ 2 kΩ
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I2 I0
- + 2 mA V0 = 4 V 3 kΩ 12 kΩ 1 kΩ
+ -
I3
- B
Fig. (b)Applying KVL around the dotted pathSum of all the voltage rise = sum of all the voltage dropVS = 12000I2 + 3000I3
HereI2 = 2 mAI3 = 4 mAVS = 12000(2 mA) + 3000(4 mA)VS = 24 + 12VS = 36 Volts
Q#2.44: In the circuit in Fig. P2.44, Vx = 6 V. Find Is.Solution:Circuit diagram:
-
Vx 2 kΩ 12 kΩ
4 kΩ+
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6 kΩ 12 kΩ
3 kΩ12 kΩ
IS
Fig. (a)Rearranging,
-
Vx 2 kΩ 12 kΩ 12 kΩ
4 kΩ+
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3 kΩ 6 kΩ12 kΩ
IS
Fig. (b)Parallel combination
3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ
18 k × k= 9 kΩ
= 2 kΩ
Parallel combination
12 kΩ × 12 kΩ= 12 kΩ + 12 kΩ
144 k × k= 24 kΩ
= 6 kΩ
-
Vx 2 kΩ 6 kΩ
4 kΩ+
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2 kΩ12 kΩ
IS
Fig. (c)Series combination= 6 kΩ + 12 kΩ= 18 kΩ
-
Vx 2 kΩ 18 kΩ
4 kΩ+
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2 kΩ
IS
Fig. (d)Parallel combination
18 kΩ × 4 kΩ= 18 kΩ + 4 kΩ
72 k × k= 22 kΩ
= 3.273 kΩ
-
Vx 2 kΩ 3.273 kΩ
+
Ix
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2 kΩ
IS
Fig. (e)According to ohm’s Law
Vx
Ix = 2 kΩ
6 VIx = 2 kΩ
Ix = 3 mA
According to current divider rule:
3.273 kΩIx = × IS
3.273 kΩ + 2 kΩ + 2 kΩ
3.273 kΩIx = × IS
7.273 kΩ
Ix = 0.451 IS
HereIx = 3 mA
IS = 6.652 mA
Q#2.28: In the network in Fig. P2.28, if Vx = 12 V, find Vs. Solution:Circuit diagram:
+ - + -
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+4 kΩ 6 kΩ
VS 6 V VX
I
2 kΩ - + -
Fig. (a) + -
+ 6 kΩ
6 V VX
I
-
Fig. (b)According to KVLSum of all the voltage rise = sum of all the voltage dropVX = 6 + 6000IHereVX = 12 V12 = 6 + 6000I6000I = 6
I = 1 mA
From fig. (a)According to KVLSum of all the voltage rise = sum of all the voltage dropVS = 4000I + 6000I + 2000I + 6
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VS = 12000I + 6HereI = 1 mAVS = 12000(0.001) + 6VS = 12 + 6
VS = 18 V
Q#2.31: If VA = 12 V in the circuit in Fig. P2.31, find VS.Solution:Circuit diagram: + VX - + -
+2 kΩ 4 kΩ
VS VA 2 VX
I
6 kΩ - - +
Fig. (a)According to KVLSum of all the voltage rise = sum of all the voltage dropVA = 4000I + 2VX + 6000IHereVX = 2000I, VA = 12 V12 = 4000I + 2(2000I) + 6000I12 = 4000I + 4000I + 6000I12 = 14000I
I = 0.857 mA
+ VX - + -
2 kΩ 4 kΩ
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VS 2 VX
I
6 kΩ - +
Fig. (b)According to KVLSum of all the voltage rise = sum of all the voltage dropVS = 2000I + 4000I + 2VX + 6000IHereVX = 2000IVS = 2000I + 4000I + 2(2000I) + 6000IVS = 2000I + 4000I + 4000I + 6000IVS = 16000I HereI = 0.857 mAVS = 16000(0.857 mA)
VS = 13.712 V
Q#2.32: A commercial power supply is modeled by the network shown in fig. P2.32.
(a) Plot V0 versus Rload for 1 Ω ≤ Rload ≤ ∞(b) What is the maximum value of V0 in (a)?(c) What is the minimum value of V0 in (a)?(d) If for some reason the output should become short circuited, that is,
Rload → 0, what current is drawn from the supply?(e) What value of Rload corresponds to maximum power consumed?
Solution:Circuit diagram:
+ 0.1 Ω
12 V V0 Rload
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-
Fig. (a)(a)
+ 0.1 Ω
12 V V0 Rload
-
Fig. (b)According to voltage division rule:
Rload
V0 = × 12 V Rload + 0.1
Rload = 1 Ω Rload = ∞ Ω
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1 ΩV0 = × 12 V 1 Ω + 0.1 Ω 1 ΩV0 = × 12 V
1.1 Ω
V0 = 10.909 Volts
Rload
V0 = × 12 V Rload + 0.1
d Rload × 12 V dRload
V0 = Lim Rload → ∞ d d Rload + 0.1 dRload dRload
Hint:V0 = 12 Volts dx
Lim = 1
x→ ∞ dx
PLOTV0 versus Rload
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10.909
12
10.2
10.4
10.6
10.8
11
11.2
11.4
11.6
11.8
12
12.2
resistance
vo
lta
ge
Series1
Series1 10.909 12
1 ¥
(b)
V0(max) = 12 V
(c)
V0(min) = 10.909 V
(d)
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+ 0.1 Ω
12 V I V0 Rload = 0 Ω
-
Fig. (c)According to ohm’s Law
12 VI = Rload + 0.1
HereRload = 0 Ω
12 VI = 0 Ω + 0.1 Ω
I = 120 A
(e)
Rload = 0.1 Ω
Q#2.33: A commercial power supply is guaranteed by the manufacturer to deliver 5 V ± 1% across a load range of 0 to 10 A. Using the circuit in Fig. P2.33 to model the supply, determine the appropriate values of R and V.Solution:Circuit diagram:
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+ R
V V0
-
Fig. (a) + -
+ R
V V0 IL
I
-
Fig. (b)
ILoad = 0 A ILoad = 10 AAccording to KVLSum of all the voltage rise = sum of all the voltage dropV = ILoadR + V0
HereV0 = 5(1 + 0.01)V0 = 5(1.01) = 5.05 VV = (0 A)R + 5.05 V5.05 V = V
According to KVLSum of all the voltage rise = sum of all the voltage dropV = IR + V0
HereV0 = V – ILoadR 5(1 – 0.01) = 5.05 V – (10 A)R5(0.99) = 5.05 V – 10R4.95 = 5.05 – 10R4.95 - 5.05 = –10R-10R = -0.1
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R = 0.01 Ω
Q#2.34: A power supply is specified to provide 48 ± 2 V at 0-200 A and is modeled by the circuit in Fig. P2.34.
(a) What are the appropriate values for V and R?(b) What is the maximum power the supply can deliver?(c) What values of Iload and V0 correspond to that level?
Solution:Circuit diagram:
+ -
+ R
V V0 ILoad
I
-
Fig. (a)(a)
ILoad = 0 A ILoad = 200 AAccording to KVLSum of all the voltage rise = sum of all the voltage dropV = ILoadR + V0
HereV0 = 48 + 2 V0 = 50 VV = (0 A)R + 50 V50 V = V
According to KVLSum of all the voltage rise = sum of all the voltage dropV = IR + V0
HereV0 = V – ILoadR 48 V - 2 V = 50 V – (200 A)R46 V = 50 V – 200R-4 V = -200RR = 0.02 Ω
(b)PLoad = ILoadV0
ILoad = 200 AV0 = V - ILoadRV0 = 50 – (200)(0.02)
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V0 = 50 – 4V0 = 46 VPLoad = (200 A)(46 V)
PLoad = 9.2 kwatt = power absorbed by the load = power delivered by the supplyILoad = 200 A & V0 = 46 V correspond to that level.
Q#2.35: Although power supply loads are often modeled as either resistors or constant current sources, some loads are best modeled as constant power loads, as indicated in Fig. P2.35. Given the model shown in the figure,
(a) Write a V-I expression for a constant power load that always draws PL watts.
(b) If PL = 40 Watt, VPS = 9 V and I0 = 5 A, determine the values of V0 and RPS.
Solution:Circuit diagram:
I0 + - + RPS
Closed path
VPS V0
-
Fig. (a)(a)
PL = VIL
(b)
PL = V0I0
PL
V0 = I0
Constt: power
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40 WV0 = 5 A
V0 = 8 VoltsApplying KVL around the closed pathSum of all the voltage rise = sum of all the voltage dropVPS = I0RPS + V0
VPS – V0
RPS = I0
9 V – 8 VRPS = 5 A
RPS = 0.2 ΩQ#2.103: Find the value of g in the network in Fig. P2.103 such that the power supplied by the 3-A source is 20 W.Solution:Circuit diagram:
IX
1 Ω 2 Ω
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3 A
gIX
2 Ω 2 Ω
Fig. (a)Using 3 A current source as a reference
- + C + -
I3 IX
2 Ω 1 Ω+
3 A
-I4 + - - + I2
A BD
2 Ω 2 Ω
gIX
Applying KCL at the Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junction
I3 + gIX = I4
Applying KCL at the Node Labeled ‘B’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionIX = I2 + gIX
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I2 = IX - gIX
I2 = IX(1 - g)
Applying KCL at the Node Labeled ‘C’Sum of all the currents entering into the junction = sum of all the currents leaving that junction3 A = IX + I3
I3 = 3 - IX
Writing KVL equation around the loop CBDCSum of all the voltage rise = sum of all the voltage dropV3A = (1)IX + 2I2
V3A = IX + 2I2
V3A =?P3A = V3AIP3A = 20 WattI = 3 A
P3A
V3A =
I
20 WattV3A =
3 A
V3A = 6.667 Volts
Therefore6.667 = IX + 2IX(1 - g)6.667 = IX + 2IX - 2gIX
6.667 = 3IX - 2gIX
6.667 = (3 - 2g)IX … (i)
Writing KVL equation around the loop CADCSum of all the voltage rise = sum of all the voltage dropV3A = 2I3 + 2I4
Substituting the corresponding values6.667 = 2(3 - IX) + 2(I3 + gIX)6.667 = 6 - 2IX + 2I3 + 2gIX
6.667 = 6 - 2IX + 2(3 - IX) + 2gIX
6.667 = 6 - 2IX + 6 - 2IX + 2gIX
6.667 = 12 - 4IX + 2gIX
6.667 - 12 = -4IX + 2gIX
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-5.333 = 2IX(-2 + g)
-2.667IX = (g - 2)Substituting the value of IX in equation (i)
-2.6676.667 = (3 - 2g)
(g - 2)
6.667(g - 2) = -2.667(3 - 2g)6.667g – 13.334 = -8.001 + 5.334g6.667g – 13.334 + 8.001 - 5.334g = 01.333g – 5.333 = 01.333g = 5.333
g = 4.001
Q#2.110: Find V0 in the circuit in Fig. P2.110.Solution:Circuit diagram:
IS
+ 3 kΩ 2000IS
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12 V IS 1 kΩ V0
-
Fig. (a)According to KVL Sum of all the voltage rise = sum of all the voltage drop12 + 2000IS = 3000IS + 1000IS
12 + 2000IS - 3000IS - 1000IS = 012 - 2000IS = 0-2000IS = -12
IS = 6 mA
According to ohm’s LawV0 = IS(1 kΩ )V0 = (6 mA)(1 kΩ )V0 = (6 × 10-3)(1 × 10+3)V0 = 6 × 10-3+3
V0 = 6 × 100
V0 = 6 × 1
V0 = 6 Volts
Q#2.99: Given V0 in the network in Fig. P2.99, find IA.Solution:Circuit diagram:
EI1
I4
- + IA
1 kΩ 1 kΩ
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+ 12 V -
I2 + - D
B 1 kΩ + +
2 kΩ 1 kΩ V0 = 4 V
6 V -
I3 I0
IX C -
A Fig. (a)According to ohm’s Law
V0
I0 = 1 kΩ
4 VI0 = 1 kΩ
I0 = 4 mA
Applying KVL around the closed pathSum of all the voltage rise = sum of all the voltage drop2000I3 = 12 V + V0
HereV0 = 4 V2000I3 = 12 V + 4 V2000I3 = 16 V
I3 = 8 mA
Applying KCL at the Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI0 + I3 = IX
Here
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I0 = 4 mAI3 = 8 mA4 mA + 8 mA = IX
IX = 12 mA
Applying KVL around the circular pathSum of all the voltage rise = sum of all the voltage drop6 V = 1000I2 + 2000I3
HereI3 = 8 mA6 V = 1000I2 + 2000(8 mA)6 V = 1000I2 + 16 V6 V – 16 V = 1000I2
-10 V = 1000I2
I2 = -10 mA
Applying KCL at the Node Labeled ‘B’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionIX = I1 + I2
I1 = IX – I2
HereIX = 12 mAI2 = -10 mAI1 = 12 mA – (-10 mA)
I1 = 22 mA
Applying KVL around the loop BEDCABSum of all the voltage rise = sum of all the voltage drop6 = 1000I1 + 1000I4 + 1000I0 HereI1 = 22 mAI0 = 4 mA6 = 1000(22 mA) + 1000I4 + 1000(4 mA) 6 = 22 + 1000I4 + 46 = 26 + 1000I4 -20 = 1000I4
I4 = -20 mA
Applying KCL at the Node Labeled ‘E’Sum of all the currents entering into the junction = sum of all the currents leaving that junction
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I1 = IA + I4
IA = I1 – I4
HereI1 = 22 mAI4 = -20 mAIA = 22 mA – (-20 mA)
IA = 42 mA
Q#2.72: Find the value of VS in the network in Fig. P2.72 such that the power supplied by the current source is 0. Solution:Circuit diagram:
3 Ω A 8 Ω
I1 I2
18 V 3 A VS
B 2 Ω 6 Ω
Power supplied by the current source = VI = 0 wattsHereI = 3 AThereforeV = 0 VoltsHence,VA = VB = 0 VApplying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI1 + I2 + 3 A = 0According to ohm’s Law
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18 VI1 = Hint: VA = VB = 0 V 2 Ω + 3 Ω
I1 = 3.6 A
According to ohm’s Law
VS
I2 = Hint: VA = VB = 0 V 6 Ω + 8 Ω
I2 = 0.071VS A
Substituting the corresponding values3.6 A + 0.071VS A + 3 A = 00.071VS = -6.6
VS = -92.958 V
Q#2.55: Find the equivalent resistance, Req, in the network in Fig. P2.55.Solution:Circuit diagram:
12 Ω
12 Ω
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12 Ω 12 Ω
Req
12 Ω 12 Ω 12 Ω
Fig. (a)
12 Ω
12 Ω
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12 Ω 12 Ω
Req
12 Ω 12 Ω 12 Ω
Fig. (b)Parallel combination
12 Ω × 12 Ω= 12 Ω + 12 Ω
144 Ω × Ω= 24 Ω
= 6 Ω
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12 Ω 12 Ω
Req
6 Ω 12 Ω
Fig. (c)
12 Ω
Req
6 Ω 12 Ω
Fig. (d)
Series combination:= 6 Ω + 12 Ω= 18 Ω
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12 Ω
Req
18 Ω
Fig. (e)Parallel combination
12 Ω × 18 Ω= 12 Ω + 18 Ω
216 Ω × Ω= 30 Ω
= 7.2 Ω = Req
Q#2.73: Find V0 in the circuit in Fig. P2.73.Solution:Circuit diagram:
4 kΩ 3 kΩ
+
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6 m V0
2 kΩ 6 kΩ 3 kΩ
-
Fig. (a)Series combination:= 2 kΩ + 4 kΩ= 6 kΩ
6 kΩ 3 kΩ
+
6 m V0
6 kΩ 3 kΩ
-
Fig. (b)Parallel combination
6 kΩ × 6 kΩ= 6 kΩ + 6 kΩ
36 k × k= 12 k
= 3 kΩ
3 kΩ 3 kΩ
+
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6 m V0
3 kΩ
-
Fig. (c)Series combination:= 3 kΩ + 3 kΩ= 6 kΩ
6 kΩ
+
6 m V0
3 kΩ
-
Fig. (d)Parallel combination
6 kΩ × 3 kΩ= 6 kΩ + 3 kΩ
18 k × k= 9 k
= 2 kΩ
2 kΩ
+
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6 m V0
-
Fig. (e)According to ohm’s Law:V0 = (2 k)(6 mA)V0 = (2 × 10+3)(6 × 10-3)V0 = 12 × 10+3-3
V0 = 12 × 100
V0 = 12 × 1V0 = -12 Volts based on polarity
Q#2.67: Determine V0 in the network in Fig. P2.67.Solution:Circuit diagram:
4 kΩ 8 kΩ
+
4 m V0
14 kΩ 9 kΩ 6 kΩ
-
Fig. (a)Series combination:= 4 kΩ + 14 kΩ= 18 kΩ
8 kΩ
+
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4 m V0
18 kΩ 9 kΩ 6 kΩ
-
Fig. (b)Parallel combination:LetR1 = 18 kΩR2 = 9 kΩR3 = 6 kΩ
R1R2R3
= R2R3 + R1R3 + R1R2
(18 kΩ )(9 kΩ )(6 kΩ )
= (9 kΩ )(6 kΩ ) + (18 kΩ )(6 kΩ ) + (18 kΩ )(9 kΩ )
972 × k × k × k =
54 × k × k + 108 × k × k + 162 × k × k
972 × k × k × k = 324 × k × k
= 3 kΩ
8 kΩ
+
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4 m V0
3 kΩ
-
Fig. (c)According to ohm’s Law:V0 = (3 k + 8 k)(4 mA)V0 = (11 × 10+3)(4 × 10-3)V0 = 44 × 10+3-3
V0 = 44 × 100
V0 = 44 × 1V0 = 44 Volts
Q#2.60: Find Vab and Vdc in the circuit in Fig. P2.60.Solution:Circuit diagram:
a + Vab - b
- + e + - 2 Ω 5 Ω
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2 A 2 A+ 20 V +
4 Ω 3 Ω
+ d- -
4 A -
Vdc
1 Ω
+ 2 Ω - +-
c f Fig. (a)
2 Ω 5 Ω
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20 V
4 Ω 3 Ω
1 Ω
2 Ω
Fig. (b)Series combinations= 3 Ω + 5 Ω= 8 Ω
= 2 Ω + 4 Ω + 2 Ω= 8 Ω
A
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20 V
8 Ω 8 Ω
1 Ω
B Fig. (c)Parallel combination
8 Ω × 8 Ω= 8 Ω + 8 Ω
64 Ω × Ω= 16 Ω
= 4 Ω
20 V
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4 Ω
I
1 Ω
Fig. (d)According to ohm’s Law
20 VI = 5 Ω
I = 4 A
According to ohm’s Law:
V4Ω = (4 Ω )(4 A)V4Ω = 16 Volts = VAB
From figure (a) According to KVL Sum of all the voltage drop = sum of all the voltage riseVab + (2 A)(2 Ω ) = (2 A)(5 Ω ) Vab + 4 V = 10 V
Vab = 6 VAccording to KVL Sum of all the voltage drop = sum of all the voltage riseVdc + (2 A)(2 Ω ) + (2 A)(2 Ω ) = 0Vdc + 4 V + 4 V = 0
Vdc = -8 V
Q#2.61: Find I1 and V0 in the circuit in Fig. P2.61.Solution:Circuit diagram:
A
I1 2 kΩ +
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6 V 12 kΩ 4 kΩ V0
-
B Fig. (a)
According to ohm’s Law
VAB
I1 = 12 kΩ
6 VI1 = 12 kΩ
I1 = 0.5 mA
According to voltage divider rule:
4 kΩV0 = × VAB
2 kΩ + 4 kΩ
4 kΩV0 = × 6 V 6 kΩ
V0 = 4 VQ#2.62: Find I1 and V0 in the circuit in Fig. P2.62.Solution:Circuit diagram:
A
2 kΩ 8 kΩ +
12 V 6 kΩ 4 kΩ V0
I1 -
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B Fig. (a)
A
2 kΩ 8 kΩ
12 V 6 kΩ 4 kΩ
I1
B Fig. (b)Series combination:= 4 kΩ + 8 kΩ= 12 kΩ
A
2 kΩ 12 kΩ
12 V 6 kΩ
I1
B Fig. (c)Parallel combination
6 kΩ × 12 kΩ= 6 kΩ + 12 kΩ
72 k × k= 18 k
= 4 kΩ
A
2 kΩ 4 kΩ
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12 V
B Fig. (c)
According to voltage divider rule:
4 kΩVAB = × 12 V 2 kΩ + 4 kΩ
4 kΩVAB = × 12 V 6 kΩ
VAB = 8 V
According to ohm’s Law
VAB
I1 = 6 kΩ
8 VI1 = 6 kΩ
I1 = 1.334 mA
According to voltage divider rule:
4 kΩV0 = × VAB
4 kΩ + 8 kΩ
4 kΩV0 = × 8 V 12 kΩ
V0 = 2.667 V
Q#2.63: Find I0 in the network in Fig. P2.63.
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Solution:Circuit diagram:
I0 6 kΩ
12 mA 12 kΩ 12 kΩ 12 kΩ
Fig. (a)Parallel combination
12 kΩ × 12 kΩ= 12 kΩ + 12 kΩ
144 k × k= 24 k
= 6 kΩ
I0 6 kΩ
12 mA 12 kΩ 6 kΩ
Fig. (b)
According to current divider rule:
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12 kΩI0 = × 12 mA 12 kΩ + 12 kΩ
12 kΩI0 = × 12 mA 24 kΩ
I0 = 6 mA
Q#2.64: Find I1 in the circuit in Fig. P2.64.Solution:Circuit diagram:
2 kΩ I1 2 kΩ
6 mA 10 kΩ 2 kΩ 2 kΩ
Fig. (a)Series combination:= 2 kΩ + 2 kΩ= 4 kΩ
A
2 kΩ I1 4 kΩ
6 mA 10 kΩ 2 kΩ
B Fig. (b)Parallel combination
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2 kΩ × 4 kΩ= 2 kΩ + 4 kΩ
8 k × k= 6 k
= 1.334 kΩ
I2
I1 2 kΩ 1.334 kΩ
6 mA 10 kΩ
Fig. (c)According to current divider rule:
10 kΩI2 = × 6 mA 3.334 kΩ + 10 kΩ
10 kΩI2 = × 6 mA 13.334 kΩ
I2 = 4.5 mAAccording to ohm’s Law:V1.334k = (1.334 k)I2
V1.334k = (1.334 k)(4.5 mA)V1.334k = (1.334 × 10+3)(4.5 × 10-3)V1.334k = 6.003 × 10+3-3
V1.334k = 6.003 × 100
V1.334k = 6.003 × 1V1.334k = 6.003 Volts = VAB
from fig. (b)According to ohm’s Law
VAB
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I1 = 2 kΩ
6.003 VI1 = 2 kΩ
I1 = -3.002 mA based on direction
Q#2.65: Determine V0 in the network in Fig. P2.65.Solution:Circuit diagram:
5 kΩ
18 mA 3 kΩ 30 mA+
1 kΩ V0
-
Fig. (a)
18 mA 30 mA 12 mA
Fig. (b)
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5 kΩ
12 mA 3 kΩ +
1 kΩ V0
I1 I2 -
Fig. (c)According to current divider rule:
3 kΩI2 = × 12 mA 3 kΩ + 6 kΩ
3 kΩI2 = × 12 mA 9 kΩ
I2 = 4 mA
According to ohm’s Law:V1k = (1 k)I2
V1k = (1 k)(4 mA)V1k = (1 × 10+3)(4 × 10-3)V1k = 4 × 10+3-3
V1k = 4 × 100
V1k = 4 × 1V1k = -4 Volts = V0 based on polarity
Q#2.66: Determine I0 in the circuit in Fig. P2.66.Solution:Circuit diagram:
12 kΩ 4 kΩ 6 kΩ
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4 kΩ 16 kΩ 2 kΩ
12 V I0
Fig. (a)Series combination:= 4 kΩ + 12 kΩ= 16 kΩ
16 kΩ 4 kΩ 6 kΩ
16 kΩ 2 kΩ
12 V I0
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Fig. (b)Parallel combination
16 kΩ × 16 kΩ= 16 kΩ + 16 kΩ
256 k × k= 32 k
= 8 kΩ
8 kΩ 4 kΩ 6 kΩ
2 kΩ
12 V I0
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Fig. (c)Series combination:= 4 kΩ + 8 kΩ= 12 kΩ 12 kΩ 6 kΩ
A
2 kΩ
12 V I0
B Fig. (d)Parallel combination
12 kΩ × 6 kΩ= 12 kΩ + 6 kΩ
72 k × k= 18 k
= 4 kΩ
4 kΩ
+ - +
2 kΩ
-
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I12 V
Fig. (e)Applying KVL around the closed pathSum of all the voltage rise = sum of all the voltage drop12 = 4000I + 2000I12 = 6000I
I = 2 mA
According to ohm’s Law:V4k = (4 k)IV4k = (4 k)(2 mA)V4k = (4 × 10+3)(2 × 10-3)V4k = 8 × 10+3-3
V4k = 8 × 100
V4k = 8 × 1V4k = 8 Volts = VAB
from fig. (d)According to ohm’s Law
VAB
I0 = 6 kΩ
8 VI0 = 6 kΩ
I0 = -1.334 mA based on direction
Q#2.68: Find I0 in the circuit in Fig. P2.68.Solution:Circuit diagram:
A 4 kΩ
9 kΩ
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6 kΩ
24 V
6 kΩ 3 kΩI0
B Fig. (a)Parallel combination
3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ
18 k × k= 9 k
= 2 kΩ
A 4 kΩ
9 kΩ
24 V
6 kΩ 2 kΩ
B Fig. (b)Series combination:= 2 kΩ + 4 kΩ= 6 kΩ A 6 kΩ
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9 kΩ
24 V
6 kΩ
B Fig. (c)Parallel combination
6 kΩ × 6 kΩ= 6 kΩ + 6 kΩ
36 k × k= 12 k
= 3 kΩ
A
9 kΩ 3 kΩ
I 24 V
B Fig. (d)Applying KVL around the closed pathSum of all the voltage rise = sum of all the voltage drop24 = 9000I + 3000I24 = 12000I
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I = 2 mA
According to ohm’s Law:V3k = (3 k)IV3k = (3 k)(2 mA)V3k = (3 × 10+3)(2 × 10-3)V3k = 6 × 10+3-3
V3k = 6 × 100
V3k = 6 × 1V3k = 6 Volts = VAB
From fig. (b)According to voltage divider rule:
2 kΩV2k = × VAB
2 kΩ + 4 kΩ
Here VAB = 6 V
2 kΩV2k = × 6 V 6 kΩ
V2k = 2 V
According to ohm’s Law
V2k
I0 = 6 kΩ
2 VI0 = 6 kΩ
I0 = 0.334 mA
Q#2.74: Find I0 in the network in Fig. P2.74.Solution:Circuit diagram:
1 kΩ 3 kΩ 2 kΩ
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3 kΩ 6 V
6 kΩ 12 kΩ 6 kΩ 3 kΩ
I0
Fig. (a)Parallel combination
3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ
18 k × k= 9 k
= 2 kΩ
1 kΩ 3 kΩ 2 kΩ
6 V
6 kΩ 12 kΩ 2 kΩ 3 kΩ
I0
Fig. (b)Series combination:= 2 kΩ + 2 kΩ= 4 kΩ
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1 kΩ 3 kΩ 4 kΩ
6 V
6 kΩ 12 kΩ 3 kΩ
I0
Fig. (c)Parallel combination
12 kΩ × 4 kΩ= 12 kΩ + 4 kΩ
48 k × k= 16 k
= 3 kΩ
1 kΩ 3 kΩ 3 kΩ
6 V
6 kΩ 3 kΩ
I0
Fig. (d)Series combination:= 3 kΩ + 3 kΩ= 6 kΩ
A
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1 kΩ 6 kΩ
6 V
6 kΩ 3 kΩ
I0
B Fig. (e)Parallel combination:LetR1 = 3 kΩR2 = 6 kΩR3 = 6 kΩ
R1R2R3
= R2R3 + R1R3 + R1R2
(3 kΩ )(6 kΩ )(6 kΩ )
= (6 kΩ )(6 kΩ ) + (3 kΩ )(6 kΩ ) + (3 kΩ )(6 kΩ )
108 × k × k × k =
36 × k × k + 18 × k × k + 18 × k × k
108 × k × k × k = 72 × k × k
= 1.5 kΩ
A 1 kΩ 1.5 kΩ
6 V I
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B Fig. (f)According to KVLSum of all the voltage rise = sum of all the voltage drop6 = 1000I + 1500I6 = 2500I
I = 2.4 mA
According to ohm’s Law:V1.5k = (1.5 k)IV1.5k = (1.5 k)(2.4 mA)V1.5k = (1.5 × 10+3)(2.4 × 10-3)V1.5k = 1.5 × 10+3-3
V1.5k = 1.5 × 100
V1.5k = 1.5 × 1V1.5k = 3.6 Volts = VAB
From fig. (f)According to ohm’s Law
VAB
I0 = 3 kΩ
3.6 VI0 = 3 kΩ
I0 = 1.2 mA
Q#2.75: Find I0 in the circuit in Fig. P2.75.Solution:Circuit diagram:
6 kΩ
6 kΩ
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12 V 3 kΩ
2 kΩ 4 kΩ
I0
Fig. (a)
6 kΩ
6 kΩ 3 kΩ
12 V
2 kΩ 4 kΩ
I0
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Fig. (b)Parallel combination
3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ
18 k × k= 9 k
= 2 kΩ
6 kΩ
2 kΩ
12 V
2 kΩ 4 kΩ
I0
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Fig. (b)Series combination:= 2 kΩ + 2 kΩ= 4 kΩ
A
6 kΩ
4 kΩ4 kΩ
12 V
I0
BFig. (c)
Parallel combination
4 kΩ × 4 kΩ= 4 kΩ + 4 kΩ
16 k × k= 8 k
= 2 kΩ
A
6 kΩ
2 kΩI
12 V
BFig. (d)
According to KVL
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Sum of all the voltage rise = sum of all the voltage drop12 = 6000I + 2000I12 = 8000I
I = 1.5 mA
According to ohm’s Law:V2k = (2 k)IV2k = (2 k)(1.5 mA)V2k = (2 × 10+3)(1.5 × 10-3)V2k = 3 × 10+3-3
V2k = 3 × 100
V2k = 3 × 1V2k = 3 Volts = VAB
From fig. (c)According to ohm’s Law
VAB
I0 = 4 kΩ
3 VI0 = 4 kΩ
I0 = 0.75 mA
Q#2.76: Determine V0 in the circuit in Fig. P2.76.Solution:Circuit diagram:
3 kΩ
6 kΩ 6 kΩ
12 V A
8 kΩ +
6 kΩ 12 kΩ 4 kΩ
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V0
-
B Fig. (a)Series combination:= 4 kΩ + 8 kΩ= 12 kΩ
3 kΩ
6 kΩ 6 kΩ
12 V A
12 kΩ
6 kΩ 12 kΩ
B Fig. (b)
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Parallel combination
12 kΩ × 12 kΩ= 12 kΩ + 12 kΩ
144 k × k= 24 k
= 6 kΩ
3 kΩ
6 kΩ 6 kΩ
12 V A
6 kΩ 6 kΩ
B Fig. (c)
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3 kΩ
6 kΩ 6 kΩ
12 V A
6 kΩ 6 kΩ
B Fig. (d)
Parallel combinations:
6 kΩ × 6 kΩ= 6 kΩ + 6 kΩ
36 k × k= 12 k
= 3 kΩ
6 kΩ × 6 kΩ= 6 kΩ + 6 kΩ
36 k × k= 12 k
= 3 kΩ
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3 kΩ
3 kΩ
12 V
A
3 kΩ
I
B Fig. (e)According to KVLSum of all the voltage rise = sum of all the voltage drop12 = 3000I + 3000I + 3000I12 = 9000I
I = 1.334 mA
According to ohm’s Law:VAB = (3 k)IVAB = (3 k)(1.334 mA)VAB = (3 × 10+3)(1.334 × 10-3)VAB = 4.002 × 10+3-3
VAB = 4.002 × 100
VAB = 4.002 × 1VAB = 4.002 Volts
From fig. (a)According to voltage divider rule:
4 kΩV0 = × VAB
4 kΩ + 8 kΩ
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Here VAB = 4.002 V
4 kΩV0 = × 4.002 V 12 kΩ
V0 = 1.334 V Answer
Q#2.77: Find V0 in the circuit in Fig. P2.77.Solution:Circuit diagram:
6 kΩ 3 kΩ 2 kΩ
8 kΩ +
12 mA 4 kΩ V0
-
Fig. (a)
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Parallel combination
3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ
18 k × k= 9 k
= 2 kΩ
2 kΩ 2 kΩ
8 kΩ -
12 mA 4 kΩ
+I1 I2
Fig. (b)
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According to current divider rule:
2 kΩ + 2 kΩI2 = × 12 mA 4 kΩ + 8 kΩ + 2 kΩ + 2 kΩ
4 kΩ I2 = × 12 mA 16 kΩ
I2 = 3 mA
According to ohm’s Law:V4k = (4 k)I2
V4k = (4 k)(3 mA)V4k = (4 × 10+3)(3 × 10-3)V4k = 12 × 10+3-3
V4k = 12 × 100
V4k = 12 × 1V4k = -12 Volts = V0 based on polarity
Q#2.78: Find V0 in the circuit in Fig. P2.78.Solution:Circuit diagram:
9 kΩ 4 kΩ 18 kΩ
6 kΩ
+ 12 mA
3 kΩ V0
-
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Fig. (a)
9 kΩ 4 kΩ 18 kΩ
12 mA
3 kΩ 6 kΩ
Fig. (b)
Parallel combination
3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ
18 k × k= 9 k
= 2 kΩ
9 kΩ 4 kΩ 18 kΩ
12 mA
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2 kΩ
Fig. (c)
Parallel combination
18 kΩ × 9 kΩ= 18 kΩ + 9 kΩ
162 k × k= 27 k
= 6 kΩ
I1 I2
6 kΩ 4 kΩ
+ 12 mA
2 kΩ V0
-Fig. (d)
According to current divider rule:
6 kΩ I2 = × 12 mA
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2 kΩ + 4 kΩ + 6 kΩ
6 kΩ I2 = × 12 mA 12 kΩ
I2 = 6 mA
According to ohm’s Law:V2k = (2 k)I2
V2k = (2 k)(4 mA)V2k = (2 × 10+3)(6 × 10-3)V2k = 12 × 10+3-3
V2k = 12 × 100
V2k = 12 × 1V2k = 12 Volts = V0 based on polarity
Q#2.79: Find I0 in the circuit in Fig. P2.79.Solution:Circuit diagram:
12 kΩ 2 kΩ
6 mA I0 12 mA 6 kΩ
12 kΩ
Fig. (a)
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12 kΩ 2 kΩ
6 mA 12 mA
6 kΩ 12 kΩ
I0
Fig. (b)
6 mA 12 mA 6 mA
Fig. (c)
12 kΩ 2 kΩ
B6 mA
6 kΩ 12 kΩ
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I0
A Fig. (d)Parallel combination
12 kΩ × 6 kΩ= 12 kΩ + 6 kΩ
72 k × k= 18 k
= 4 kΩ
12 kΩ 2 kΩ
B6 mA
4 kΩ
I1 I2
AFig. (e)
According to current divider rule:
12 kΩ I2 = × 6 mA
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12 kΩ + 2 kΩ + 4 kΩ
12 kΩ I2 = × 6 mA 18 kΩ
I2 = 4 mA
According to ohm’s Law:V4k = (4 k)I2
V4k = (4 k)(4 mA)V4k = (4 × 10+3)(4 × 10-3)V4k = 16 × 10+3-3
V4k = 16 × 100
V4k = 16 × 1V4k = 16 Volts = VAB
From fig. (d)According to ohm’s Law
VAB
I0 = 6 kΩ
16 VI0 = 6 kΩ
I0 = -2.667 mA based on direction
Q#2.80: Find I0 in the circuit in Fig. P2.80.Solution:Circuit diagram:
9 kΩ 4 kΩ 8 kΩ 18 kΩ
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12 V
4 kΩ
I0
Fig. (a)
9 kΩ 4 kΩ 8 kΩ 18 kΩ
4 kΩ 12 V
I0
Fig. (b)Parallel combination:LetR1 = 9 kΩR2 = 4 kΩR3 = 18 kΩ
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R1R2R3
= R2R3 + R1R3 + R1R2
(9 kΩ )(4 kΩ )(18 kΩ )
= (4 kΩ )(18 kΩ ) + (9 kΩ )(18 kΩ ) + (9 kΩ )(4 kΩ )
648 × k × k × k =
72 × k × k + 162 × k × k + 36 × k × k
648 × k × k × k = 270 × k × k
= 2.4 kΩ
2.4 kΩ 8 kΩ
4 kΩ 12 V
I0
Fig. (c)Series combination:= 2.4 kΩ + 8 kΩ= 10.4 kΩ
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10.4 kΩ
B
4 kΩ 12 V
I0
A Fig. (d)According to ohm’s Law
VAB
I0 = 4 kΩ
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12 VI0 = 4 kΩ
I0 = -3 mA based on direction
Q#2.81: Find I0 in the circuit in Fig. P2.81.Solution:Circuit diagram:
3 kΩ
9 kΩ 6 kΩ 12 kΩ
12 V
12 kΩ
4 kΩ
I0
Fig. (a)
3 kΩ
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9 kΩ 6 kΩ 12 kΩ
12 V
4 kΩ 12 kΩ
I0
Fig. (b)
Parallel combination
12 kΩ × 4 kΩ= 12 kΩ + 4 kΩ
48 k × k= 16 k
= 3 kΩ
A
3 kΩ
9 kΩ 6 kΩ 12 kΩ
12 V C
3 kΩ
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B Fig. (c)Series combination:= 3 kΩ + 9 kΩ= 12 kΩ
A
3 kΩ
12 kΩ 6 kΩ 12 kΩ 12 V
B
Fig. (d)Parallel combination:LetR1 = 12 kΩR2 = 6 kΩR3 = 12 kΩ
R1R2R3
= R2R3 + R1R3 + R1R2
(12 kΩ )(6 kΩ )(12 kΩ )
= (6 kΩ )(12 kΩ ) + (12 kΩ )(12 kΩ ) + (12 kΩ )(6 kΩ )
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864 × k × k × k =
72 × k × k + 144 × k × k + 72 × k × k
864 × k × k × k = 288 × k × k
= 3 kΩ
A
3 kΩ
I 3 kΩ 12 V
B
Fig. (e)According to KVLSum of all the voltage rise = sum of all the voltage drop12 = 3000I + 3000I12 = 6000I
I = 2 mA
According to ohm’s Law:V3k = (3 k)IV3k = (3 k)(2 mA)V3k = (3 × 10+3)(2 × 10-3)V3k = 6 × 10+3-3
V3k = 6 × 100
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V3k = 6 × 1V3k = 6 Volts = VAB
From fig. (a)According to voltage divider rule:
3 kΩVCB = × VAB
3 kΩ + 9 kΩ
Here VAB = 6 V
3 kΩVCB = × 6 V 12 kΩ
VCB = 1.5 VAccording to ohm’s Law
VCB
I0 = 4 kΩ
1.5 VI0 = 4 kΩ
I0 = 0.375 mA
Q#2.82: Find V0 in the circuit in Fig. P2.82.Solution:Circuit diagram:
9 kΩ 6 kΩ 12 kΩ
+ V0 -12 V
12 kΩ
4 kΩ
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Fig. (a)
9 kΩ 6 kΩ 12 kΩ
12 V
4 kΩ 12 kΩ
Fig. (b)Parallel combination
12 kΩ × 4 kΩ= 12 kΩ + 4 kΩ
48 k × k
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= 16 k
= 3 kΩ
A
9 kΩ 6 kΩ 12 kΩ
12 V C
3 kΩ
B Fig. (c)
According to voltage divider rule:
3 kΩVCB = × VAB
3 kΩ + 9 kΩ
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Here VAB = 12 V
3 kΩVCB = × 12 V 12 kΩ
VCB = 3 V = V0
Q#2.83: Find I0 in the circuit in Fig. P2.83.Solution:Circuit diagram:
4 kΩ 6 kΩ 12 kΩ
12 mA
6 kΩ
I0
Fig. (a)
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4 kΩ 6 kΩ 12 kΩ
12 mA
6 kΩ
I0
Fig. (b)
4 kΩ 6 kΩ 12 kΩ
12 mA
Short circuit
Fig. (c)Because of short circuit
I0 = 0 A
Q#2.84: Determine the value of V0 in the circuit in Fig. P2.84.Solution:Circuit diagram:
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+6 V
V0 4 kΩ 4 kΩ 12 kΩ
- 4 kΩ
3 kΩ
6 kΩ
12 V
Fig. (a)
6 V 6 V
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12 VFig. (b)
+6 V
V0 4 kΩ 4 kΩ 12 kΩ
- 4 kΩ
3 kΩ 6 kΩ
Fig. (c)Parallel combination
3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ
18 k × k
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= 9 k
= 2 kΩ
+6 V
V0 4 kΩ 4 kΩ 12 kΩ
- 4 kΩ
2 kΩ
Fig. (d)Series combination:= 2 kΩ + 4 kΩ= 6 kΩ
A
+
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6 V
4 kΩ V0 4 kΩ 6 kΩ 12 kΩ
-
B Fig. (e)Parallel combination:LetR1 = 4 kΩR2 = 6 kΩR3 = 12 kΩ
R1R2R3
= R2R3 + R1R3 + R1R2
(4 kΩ )(6 kΩ )(12 kΩ )
= (6 kΩ )(12 kΩ ) + (4 kΩ )(12 kΩ ) + (4 kΩ )(6 kΩ )
864 × k × k × k =
72 × k × k + 48 × k × k + 24 × k × k
288 × k × k × k = 144 × k × k
= 2 kΩ
A
6 V
4 kΩ 2 kΩ I
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B Fig.(f)According to KVLSum of all the voltage rise = sum of all the voltage drop6 = 4000I + 2000I6 = 6000I
I = 1 mA
According to ohm’s Law:VAB = (2 k)IVAB = (2 k)(1 mA)VAB = (2 × 10+3)(1 × 10-3)VAB = 2 × 10+3-3
VAB = 2 × 100
VAB = 2 × 1VAB = 2 Volts = V0
Q#2.85: Find P4Ω in the network in Fig. P2.85.Solution:Circuit diagram:
12 Ω12 Ω
6 Ω
9 Ω 12 Ω
4 Ω24 V
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12 Ω 12 Ω
Fig. (a)
12 Ω 12 Ω 6 Ω
9 Ω 12 Ω
C
24 V
12 Ω 12 Ω 4 Ω
D I4Ω
Fig. (b)
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Parallel combination
12 Ω × 4 Ω= 12 Ω + 4 Ω
48 Ω × Ω= 16 Ω
= 3 Ω
Parallel combination:LetR1 = 12 ΩR2 = 6 ΩR3 = 12 Ω
R1R2R3
= R2R3 + R1R3 + R1R2
(12 Ω )(6 Ω )(12 Ω )
= (6 Ω )(12 Ω ) + (12 Ω )(12 Ω ) + (12 Ω )(6 Ω )
864 × Ω × Ω × Ω =
72 × Ω × Ω + 144 × Ω × Ω + 72 × Ω × Ω
864 × Ω × Ω × Ω = 288 × Ω × Ω
= 3 Ω
3 Ω
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9 Ω 12 Ω
C
24 V
12 Ω 3 Ω
D
Fig. (c)Series combination:= 3 Ω + 9 Ω= 12 Ω
12 Ω
24 V
12 Ω
12 Ω 3 Ω
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Fig. (d)Parallel combination
12 Ω × 12 Ω= 12 Ω + 12 Ω
144 Ω × Ω= 24 Ω
= 6 Ω
A
12 Ω
24 V
6 Ω
B
3 Ω
Fig. (d)
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According to voltage divider rule:
3 ΩV3Ω = × VBA
3 Ω + 6 Ω
Here VBA = 24 V
3 ΩV3Ω = × 24 V 9 Ω
V3Ω = 8 V = VCD
According to ohm’s Law
VCD
I4Ω = 4 Ω
8 VI4Ω = 4 Ω
I4Ω = 2 A
According to ohm’s Law:P4Ω = VCDIP4Ω = (8 V)(2 A)P4Ω = 16 watts
Q#2.86: Find I0 in the network in Fig. P2.86.Solution:Circuit diagram:
12 Ω 12 Ω
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6 Ω I0
12 Ω 6 mA
12 Ω 12 Ω 3 Ω 3 Ω
Fig. (a)
B
12 Ω 6 Ω 12 Ω
6 mAI0
A
12 Ω 12 Ω 12 Ω 3 Ω 3 Ω
Fig. (b)Parallel combination
12 Ω × 12 Ω= 12 Ω + 12 Ω
144 Ω × Ω
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= 24 Ω
= 6 Ω
Parallel combination
6 Ω × 12 Ω= 6 Ω + 12 Ω
72 Ω × Ω= 18 Ω
= 4 Ω
B
4 Ω 12 Ω
6 mA
A
6 Ω 12 Ω 3 Ω 3 Ω
Fig. (c)Parallel combination:
LetR1 = 12 ΩR2 = 6 ΩR3 = 12 Ω
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R1R2R3
= R2R3 + R1R3 + R1R2
(12 Ω )(6 Ω )(12 Ω )
= (6 Ω )(12 Ω ) + (12 Ω )(12 Ω ) + (12 Ω )(6 Ω )
864 × Ω × Ω × Ω =
72 × Ω × Ω + 144 × Ω × Ω + 72 × Ω × Ω
864 × Ω × Ω × Ω = 288 × Ω × Ω
= 3 Ω
B
4 Ω 3 Ω
6 mA
A
3 Ω 3 Ω
Fig. (d)Parallel combination
3 Ω × 3 Ω= 3 Ω + 3 Ω
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9 Ω × Ω= 6 Ω
= 1.5 Ω
B
4 Ω 1.5 Ω
6 mA
A
3 Ω
Fig. (e)Series combination:= 1.5 Ω + 3 Ω= 4.5 Ω
B
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4 Ω 4.5 Ω
6 mAI1 I2
A Fig. (f)According to current divider rule:
4.5 Ω I2 = × 6 mA 4 Ω + 4.5 Ω
4.5 Ω I2 = × 6 mA 8.5 Ω
I2 = 3.176 mA
According to ohm’s Law:VAB = (4 Ω )I2
VAB = (4 Ω )(3.176 mA)VAB = 0.013 V
From fig. (b)According to ohm’s Law
VAB
I0 = 6 Ω
0.013 VI0 = 6 Ω
I0 = 2.167 mA
Q#2.87: In the network in Fig. P2.87, the power absorbed by the 4-Ω resistor is 100 W. Find VS. Solution:Circuit diagram:
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A
3 Ω I I2 7 ΩI1
4 Ω 3 Ω
VS
B Fig. (a)P4Ω = I1
2(4 Ω )
P4Ω
I1 = 4 Ω
100 WI1 = 4 Ω
I1 = 5 A
According to ohm’s Law:VAB = (4 Ω )I1
VAB = (4 Ω )(5 A)VAB = 20 V
VAB
I2 = 3 Ω + 7 Ω
20 VI2 = 10 Ω
I2 = 2 A
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junction
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I = I1 + I2
I = 5 A + 2 A
I = 7 A
Applying KVL around the dotted pathAccording to KVLSum of all the voltage rise = sum of all the voltage dropVS = 3I + 4I1
VS = 3(7) + 4(5)VS = 21 + 20
VS = 41 Volts
Q#2.88: If V0 = 2 V in the circuit in Fig. P2.88, find VS. Solution:Circuit diagram:
A + -
9 kΩ I I0 4 kΩ + I1
6 kΩ 2 kΩ
VS V0 = 2 V
-
B C Fig. (a)According to ohm’s Law
V0
I0 = 2 kΩ
2 VI0 = 2 kΩ
I0 = 1 mA
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According to KVLSum of all the voltage rise = sum of all the voltage dropVAC = 4000I0 + 2000I0
VAC = 6000I0 VAC = 6000(1 mA)VAC = (6 × 10+3)(1 × 10-3)
VAC = 6 Volts = VAB
According to ohm’s Law
VAB
I1 = 6 kΩ
6 VI1 = 6 kΩ
I1 = 1 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI = I1 + I0
I = 1 mA + 1 mA
I = 2 mA
Applying KVL around the dotted pathAccording to KVLSum of all the voltage rise = sum of all the voltage dropVS = 9000I + 6000I1
VS = 9000(2 mA) + 6000(1 mA)VS = (9 × 10+3)(2 × 10-3) + (6 × 10+3)(1 × 10-3)VS = 18 + 6
VS = 24 Volts
Q#2.89: If V0 = 6 V in the circuit in Fig. P2.89, find IS. Solution:Circuit diagram:
A + -
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2 kΩ I1 I0 9 kΩ +
3 kΩ V0 = 6 V 4 kΩ
IS
-
B C Fig. (a)
According to ohm’s Law
V0
I0 = 3 kΩ 6 VI0 = 3 kΩ
I0 = 2 mA
According to KVLSum of all the voltage rise = sum of all the voltage dropVAC = 9000I0 + 3000I0
VAC = 12000I0 VAC = 12000(2 mA)VAC = (12 × 10+3)(2 × 10-3)
VAC = 24 Volts = VAB
According to ohm’s Law
VAB
I1 = 2 kΩ + 4 kΩ
24 VI1 = 6 kΩ
I1 = 4 mA
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Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionIS = I1 + I0
IS = 4 mA + 2 mA
IS = 6 mA
Q#2.90: If I0 = 2 mA in the circuit in Fig. P2.90, find VS.Solution:Circuit diagram:
A
1 kΩ 1I I2 I0
6 kΩ 3 kΩ 12 kΩ
VS
I3
B Fig. (a)According to ohm’s Law:VAB = (3 kΩ )I0
VAB = (3 kΩ )(2 mA)VAB = (3 × 10+3)(2 × 10-3)VAB = 6 × 10+3-3
VAB = 6 × 100
VAB = 6 × 1VAB = 6 V
VAB
I2 = 12 kΩ
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6 VI2 = 12 kΩ
I2 = 0.5 mA
VAB
I3 = 6 kΩ
6 VI3 = 6 kΩ
I3 = 1 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI1 = I2 + I0 + I3
I1 = 0.5 mA + 2 mA + 1 mA
I1 = 3.5 mA
Applying KVL around the dotted pathAccording to KVLSum of all the voltage rise = sum of all the voltage dropVS = 1000I1 + 6000I3
VS = 1000(3.5 mA) + 6000(1 mA)VS = (1 × 10+3)(3.5 × 10-3) + (6 × 10+3)(1 × 10-3)VS = 3.5 + 6
VS = 9.5 Volts
Q#2.91: If V1 = 5 V in the circuit in Fig. P2.91, find IS.Solution:Circuit diagram:
V1 = 5 V A + -
10 kΩ
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4 kΩ 6 kΩ 3 kΩ IS
B Fig. (a)Parallel combination
3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ
18 k × k= 9 k
= 2 kΩ
V1 = 5 V A + -
I1 10 kΩ I2 +
4 kΩ 2 kΩ
IS
-
B Fig. (b)
According to ohm’s Law
V1
I10k = 10 kΩ
5 VI10k = 10 kΩ
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I10k = 0.5 mA = I2
According to KVLSum of all the voltage rise = sum of all the voltage dropVAB = 10000I2 + 2000I2
VAB = 10000(0.5 mA) + 2000(0.5 mA)VAB = (10 × 10+3)(0.5 × 10-3) + (2 × 10+3)(0.5 × 10-3)VAB = 5 × 10+3-3 + 1 × 10+3-3
VAB = 5 × 100 + 1 × 100
VAB = 5 × 1 + 1 × 1VAB = 5 + 1
VAB = 6 Volts
According to ohm’s Law
VAB
I1 = 4 kΩ
6 VI1 = 4 kΩ
I1 = 1.5 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionIS = I1 + I2
IS = 1.5 mA + 0.5 mA
IS = 2 mA
Q#2.92: In the network in Fig. P2.92, V1 = 12 V. Find VS.Solution:Circuit diagram:
V1
+ -
2 kΩ 4 kΩ 1 kΩ
6 kΩ 4 kΩ 3 kΩ
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VS
Fig. (a)Series combination:= 1 kΩ + 3 kΩ= 4 kΩ
V1
+ -
2 kΩ 4 kΩ
6 kΩ 4 kΩ 4 kΩ
VS
Fig. (b)Parallel combination
4 kΩ × 4 kΩ= 4 kΩ + 4 kΩ
16 k × k= 8 k
= 2 kΩ
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V1
A + -
I1 2 kΩ 4 kΩ I2 +I3
6 kΩ 2 kΩ
VS
-
BFig. (c)
According to ohm’s Law
V1
I4k = 4 kΩ
12 VI4k = 4 kΩ
I4k = 3 mA = I2
According to KVLSum of all the voltage rise = sum of all the voltage dropVAB = 4000I2 + 2000I2
VAB = 4000(3 mA) + 2000(3 mA)VAB = (4 × 10+3)(3 × 10-3) + (2 × 10+3)(3 × 10-3)VAB = 12 × 10+3-3 + 6 × 10+3-3
VAB = 12 × 100 + 6 × 100
VAB = 12 × 1 + 6 × 1VAB = 12 + 6
VAB = 18 Volts
VAB
I3 = 6 kΩ
18 V
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I3 = 6 kΩ
I3 = 3 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI1 = I2 + I3
I1 = 3 mA + 3 mA
I1 = 6 mA
Applying KVL around the dotted pathAccording to KVLSum of all the voltage rise = sum of all the voltage dropVS = 2000I1 + 6000I3
VS = 2000(6 mA) + 6000(3 mA)VS = (2 × 10+3)(6 × 10-3) + (6 × 10+3)(3 × 10-3)VS = 12 + 18
VS = 30 Volts
Q#2.93: In the circuit in Fig. P2.93, V0 = 2 V. Find IS.Solution:Circuit diagram:
A C I4
2 Ω I1 I2 6 Ω 8 Ω + I3
10 Ω 12 Ω 4 Ω V0
IS
-
B D Fig. (a)
According to ohm’s Law
V0
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I4Ω = 4 Ω
2 VI4Ω = 4 Ω
I4Ω = 0.5 A = I4
According to KVLSum of all the voltage rise = sum of all the voltage dropVCD = 8I4Ω + 4I4Ω
VCD = 12I4Ω
VCD = 12(0.5 A)
VCD = 6 V
VCD
I3 = 12 Ω
6 VI3 = 12 Ω
I3 = 0.5 A
Applying KCL at Node Labeled ‘C’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI2 = I3 + I4
I2 = 0.5 A + 0.5 A
I2 = 1 A
According to KVLSum of all the voltage rise = sum of all the voltage dropVAD = 6I2 + 12I3
VAD = 6(1) + 12(0.5)VAD = 6 + 6
VAD = 12 Volts = VAB
VAB
I1 = 2 Ω + 10 Ω
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12 VI1 = 12 Ω
I1 = 1 A
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionIS = I1 + I2
IS = 1 A + 1 A
IS = 2 A
Q#2.94: In the network in Fig. P2.94, V0 = 6 V. Find IS.Solution:Circuit diagram:
A I2 C I4
I1 3 kΩ I3 1 kΩ +
7 kΩ 2 kΩ V0 2 kΩ
IS
-
B D Fig. (a)According to ohm’s Law
V0
I3 = 2 kΩ
6 VI3 = 2 kΩ
I3 = 3 mA
V0
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I4 = 1 kΩ + 2 kΩ
6 VI4 = 3 kΩ
I4 = 2 mA
Applying KCL at Node Labeled ‘C’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI2 = I3 + I4
I2 = 3 mA + 2 mA
I2 = 5 mA
According to KVLSum of all the voltage rise = sum of all the voltage dropVAB = VAC + VCD
VAB = 3000I2 + 2000I3
VAB = 3000(5 mA) + 2000(3 mA)VAB = (3 × 10+3)(5 × 10-3) + (2 × 10+3)(3 × 10-3)VAB = 15 × 10+3-3 + 6 × 10+3-3
VAB = 15 × 100 + 6 × 100
VAB = 15 × 1 + 6 × 1VAB = 15 + 6
VAB = 21 Volts
According to ohm’s Law
VAB
I1 = 7 kΩ
21 VI1 = 7 kΩ
I1 = 3 mA
Applying KCL at Node Labeled ‘C’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionIS = I1 + I2
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IS = 3 mA + 5 mA
IS = 8 mA
Q#2.95: If I0 = 4 mA in the circuit in Fig. P2.95, find IS.Solution:Circuit diagram:
A I2 C I3
I1 1 kΩ I0 +
10 kΩ 4 kΩ 2 kΩ
IS
-
12 V
B D Fig. (a)
According to ohm’s Law:V4k = (4 kΩ )I0
V4k = (4 kΩ )(4 mA)V4k = (4 × 10+3)(4 × 10-3)V4k = 16 × 10+3-3
V4k = 16 × 100
V4k = 16 × 1
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V4k = 16 V
VCD = V4k – 12 VVCD = 16 V – 12 VVCD = 4 V
VCD
I3 = 2 kΩ
4 VI3 = 2 kΩ
I3 = 2 mAApplying KCL at Node Labeled ‘C’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI2 = I0 + I3
I2 = 4 mA + 2 mA
I2 = 6 mA
According to KVLSum of all the voltage rise = sum of all the voltage dropVAB = VAC + VCD
VAB = 1000I2 + 4 VVAB = 1000(6 mA) + 4 VVAB = (1 × 10+3)(6 × 10-3) + 4 VVAB = 6 × 10+3-3 + 4 VVAB = 6 × 100 + 4 VVAB = 6 × 1 + 4 VVAB = 6 + 4
VAB = 10 Volts
VAB
I1 = 10 kΩ
10 VI1 = 10 kΩ
I1 = 1 mA
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Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionIS = I1 + I2
IS = 1 mA + 6 mA
IS = 7 mA
Q#2.96: If V0 = 6 V in the circuit in Fig. P2.96, find IS.Solution:Circuit diagram:
A + -
1 kΩ I1 I2 4 kΩ + 4 V
2 kΩ 6 kΩ V0 = 6 V
IS
-
B Fig. (a)
According to ohm’s Law
V0
I6k = 6 kΩ
6 VI6k = 6 kΩ
I6k = 1 mA = I2
According to KVLSum of all the voltage rise = sum of all the voltage dropVAB + 4 V = 4000I2 + 6000I2
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VAB + 4 V = 10000I2 VAB + 4 V = (10 × 10+3)(1 × 10-3)VAB + 4 V = 10 × 10+3-3
VAB + 4 V = 10 × 100
VAB + 4 V = 10 × 1VAB + 4 V = 10
VAB = 6 Volts
VAB
I1 = 1 kΩ + 2 kΩ
6 VI1 = 3 kΩ
I1 = 2 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionIS = I1 + I2
IS = 2 mA + 1 mA
IS = 3 mA
Q#2.98: Find I0 in the circuit in Fig. P2.98.Solution:Circuit diagram:
I0
4 kΩ 6 kΩ
3 kΩ 4 kΩ
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12 kΩ12 mA
Fig. (a)
I0
4 kΩ 6 kΩ 4 kΩ
3 kΩ 12 kΩ 12 mA
Fig. (b)Parallel combination:LetR1 = 4 kΩR2 = 6 kΩR3 = 12 kΩ
R1R2R3
= R2R3 + R1R3 + R1R2
(4 kΩ )(6 kΩ )(12 kΩ )
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= (6 kΩ )(12 kΩ ) + (4 kΩ )(12 kΩ ) + (4 kΩ )(6 kΩ )
864 × k × k × k =
72 × k × k + 48 × k × k + 24 × k × k
288 × k × k × k = 144 × k × k
= 2 kΩ
I0
4 kΩ
3 kΩ 2 kΩ 12 mA
Fig. (c)According to current divider rule:
3 kΩ I0 = × 12 mA 4 kΩ + 2 kΩ + 3 kΩ
3 kΩ I0 = × 12 mA 9 kΩ
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I0 = -4 mA Answer
Q#100: Given I0 = 2 mA in the circuit in Fig. P2.100, find IA.Solution:Circuit diagram:
D + -I1
1 kΩ 2 kΩ6 V
6 V - +
B C E I2
-I3 + I0 + IA
1 kΩ 2 kΩ 1 kΩ
- - + I4
A Fig. (a)Applying KVL around the dotted pathAccording to KVLSum of all the voltage rise = sum of all the voltage drop1000I3 = 6 + 2000I0
1000I3 = 6 + 2000(2 mA)1000I3 = 6 + (2 × 10+3)(2 × 10-3)1000I3 = 6 + 4 × 10+3-3
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1000I3 = 6 + 4 × 100
1000I3 = 6 + 4 × 11000I3 = 6 + 4 1000I3 = 10
I3 = 10 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents leaving the junction = sum of all the currents entering that junctionI4 = I3 + I0 I4 = 10 mA + 2 mA
I4 = 12 mA
Applying KVL around the path DBCDAccording to KVLSum of all the voltage rise = sum of all the voltage drop1000I1 = 6 + 61000I1 = 12
I1 = 12 mA
Applying KVL around the path DEACDAccording to KVLSum of all the voltage rise = sum of all the voltage drop2000I2 + 1000I4 + 2000I0 + 1000I1 = 02000I2 + 1000(12 mA) + 2000(2 mA) + 1000(12 mA) = 02000I2 + (1 × 10+3)(12 × 10-3) + (2 × 10+3)(2 × 10-3) + (1 × 10+3)(12 × 10-3) = 02000I2 + 12 × 10+3-3 + 4 × 10+3-3 + 12 × 10+3-3 = 02000I2 + 12 × 100 + 4 × 100 + 12 × 100 = 02000I2 + 12 × 1 + 4 × 1 + 12 × 1 = 02000I2 + 12 + 4 + 12 = 02000I2 + 28 = 02000I2 = -28
I2 = -14 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents leaving the junction = sum of all the currents entering that junctionI2 = IA + I4 -14 mA = IA + 12 mA
IA = -26 mA
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Q#2.101: Given I0 = 2 mA in the network in Fig. P2.101, find VA. Solution:Circuit diagram:
-
1 kΩ6 mA VA
I4
+ + -
BI3
6 V - 1 kΩ - +
1 kΩ 2 kΩ 2 kΩ
+I0 - I1 I2 +
A Fig. (a)
Applying KVL around the dotted pathAccording to KVLSum of all the voltage rise = sum of all the voltage drop6 + 2000I1 + 1000I0 = 06 + 2000I1 + 1000(2 mA) = 06 + 2000I1 + (1 × 10+3)(2 × 10-3) = 06 + 2000I1 + 2 × 10+3-3 = 06 + 2000I1 + 2 × 100 = 0
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6 + 2000I1 + 2 × 1 = 06 + 2000I1 + 2 = 02000I1 = -8
I1 = -4 mA
Applying KCL at Node Labeled ‘A’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI0 = I1 + I2 2 mA = -4 mA + I2
I2 = 6 mA
Applying KVL around the circular pathAccording to KVLSum of all the voltage rise = sum of all the voltage drop2000I2 = 2000I1 + 1000I3
2000(6 mA) = 2000(-4 mA) + 1000I3
(2 × 10+3)(6 × 10-3) = (2 × 10+3)(-4 × 10-3) + 1000I3
12 × 10+3-3 = -8 × 10+3-3 + 1000I3
12 × 100 = -8 × 100 + 1000I3
12 × 1 = -8 × 1 + 1000I3
12 = -8 + 1000I3
20 = 1000I3
I3 = 20 mA
Applying KCL at Node Labeled ‘B’Sum of all the currents entering into the junction = sum of all the currents leaving that junctionI2 + I3 = I4
6 mA + 20 mA = I4
I4 = 26 mA
Applying KVL around the right triangle pathAccording to KVLSum of all the voltage rise = sum of all the voltage drop0 = 1000I3 + 1000I4 + VA
0 = 1000(20 mA) + 1000(26 mA) + VA
0 = (1 × 10+3)(20 × 10-3) + (1 × 10+3)(26 × 10-3) + VA
0 = 20 × 10+3-3 + 26 × 10+3-3 + VA
0 = 20 × 100 + 26 × 100 + VA
0 = 20 × 1 + 26 × 1 + VA
0 = 20 + 26 + VA
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0 = 46 + VA
VA = -46 V