Basic Engineering Circuit Analysis

Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 1Cell: 0300-8454295; Tel: 042-5421893

Basic Engineering Circuit Analysis

CH#2


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!


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


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


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)


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)



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

A


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


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


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)


- +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


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 =


4 Ω

I3 = 2 A

VB V1

I4


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:


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


I1 = 6 A

VA VB

I3


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


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


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:


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


V0 – VB

I0 = 1 kΩ

HereV0 = 4 V, VB = 0 V

4 – 0I0 =

1 kΩ

4 VI0 = 1 kΩ


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


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+


6 kΩ 12 kΩ

3 kΩ12 kΩ

IS

Fig. (a)Rearranging,

-

Vx 2 kΩ 12 kΩ 12 kΩ

4 kΩ+


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


12 kΩ × 12 kΩ= 12 kΩ + 12 kΩ

144 k × k= 24 kΩ

= 6 kΩ

-

Vx 2 kΩ 6 kΩ

4 kΩ+


2 kΩ12 kΩ

IS

Fig. (c)Series combination= 6 kΩ + 12 kΩ= 18 kΩ

-

Vx 2 kΩ 18 kΩ

4 kΩ+


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


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:

+ - + -


+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


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


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


-

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


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


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)


+ 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:


+ 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


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?


+ -

+ 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)


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.


I0 + - + RPS

Closed path

VPS V0

-

Fig. (a)(a)

PL = VIL

(b)

PL = V0I0

PL

V0 = I0

Constt: power


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


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


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


-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


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


+ 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


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


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


18 VI1 = Hint: VA = VB = 0 V 2 Ω + 3 Ω

I1 = 3.6 A


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


12 Ω 12 Ω

Req

12 Ω 12 Ω 12 Ω

Fig. (a)

12 Ω

12 Ω


12 Ω 12 Ω

Req

12 Ω 12 Ω 12 Ω


12 Ω × 12 Ω= 12 Ω + 12 Ω

144 Ω × Ω= 24 Ω

= 6 Ω


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


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


4 kΩ 3 kΩ

+


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

-


6 kΩ × 6 kΩ= 6 kΩ + 6 kΩ

36 k × k= 12 k

= 3 kΩ

3 kΩ 3 kΩ

+


6 m V0

3 kΩ

-

Fig. (c)Series combination:= 3 kΩ + 3 kΩ= 6 kΩ

6 kΩ

+

6 m V0

3 kΩ

-


6 kΩ × 3 kΩ= 6 kΩ + 3 kΩ

18 k × k= 9 k

= 2 kΩ

2 kΩ

+


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

-


8 kΩ

+


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

+


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


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


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


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


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 +


6 V 12 kΩ 4 kΩ V0

-

B Fig. (a)


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 -


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


6 kΩ × 12 kΩ= 6 kΩ + 12 kΩ

72 k × k= 18 k

= 4 kΩ

A

2 kΩ 4 kΩ


12 V

B Fig. (c)


4 kΩVAB = × 12 V 2 kΩ + 4 kΩ

4 kΩVAB = × 12 V 6 kΩ

VAB = 8 V


VAB

I1 = 6 kΩ

8 VI1 = 6 kΩ

I1 = 1.334 mA


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.



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)



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


A

2 kΩ I1 4 kΩ

6 mA 10 kΩ 2 kΩ

B Fig. (b)Parallel combination


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


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)


5 kΩ

12 mA 3 kΩ +

1 kΩ V0

I1 I2 -


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


4 kΩ 16 kΩ 2 kΩ

12 V I0


16 kΩ 4 kΩ 6 kΩ

16 kΩ 2 kΩ

12 V I0



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


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

-


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



A 4 kΩ

9 kΩ


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


9 kΩ

24 V

6 kΩ


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


I = 2 mA


V3k = 6 × 100


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


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


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


I0

Fig. (b)Series combination:= 2 kΩ + 2 kΩ= 4 kΩ


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


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


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


6 kΩ

6 kΩ


12 V 3 kΩ

2 kΩ 4 kΩ

I0

Fig. (a)

6 kΩ

6 kΩ 3 kΩ

12 V

2 kΩ 4 kΩ

I0



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


Fig. (b)Series combination:= 2 kΩ + 2 kΩ= 4 kΩ

A

6 kΩ

4 kΩ4 kΩ

12 V

I0

BFig. (c)


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


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


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


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)



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)


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


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


Here VAB = 4.002 V

4 kΩV0 = × 4.002 V 12 kΩ

V0 = 1.334 V Answer


6 kΩ 3 kΩ 2 kΩ

8 kΩ +

12 mA 4 kΩ V0

-

Fig. (a)



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)



2 kΩ + 2 kΩI2 = × 12 mA 4 kΩ + 8 kΩ + 2 kΩ + 2 kΩ

4 kΩ I2 = × 12 mA 16 kΩ

I2 = 3 mA


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


9 kΩ 4 kΩ 18 kΩ

6 kΩ

+ 12 mA

3 kΩ V0

-


Fig. (a)

9 kΩ 4 kΩ 18 kΩ

12 mA

3 kΩ 6 kΩ

Fig. (b)


3 kΩ × 6 kΩ= 3 kΩ + 6 kΩ

18 k × k= 9 k

= 2 kΩ

9 kΩ 4 kΩ 18 kΩ

12 mA


2 kΩ

Fig. (c)


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)


6 kΩ I2 = × 12 mA


2 kΩ + 4 kΩ + 6 kΩ

6 kΩ I2 = × 12 mA 12 kΩ

I2 = 6 mA


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


12 kΩ 2 kΩ

6 mA I0 12 mA 6 kΩ

12 kΩ

Fig. (a)


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


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)


12 kΩ I2 = × 6 mA


12 kΩ + 2 kΩ + 4 kΩ

12 kΩ I2 = × 6 mA 18 kΩ

I2 = 4 mA


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





12 V

4 kΩ

I0

Fig. (a)


4 kΩ 12 V

I0



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


10.4 kΩ

B

4 kΩ 12 V

I0

A Fig. (d)According to ohm’s Law

VAB

I0 = 4 kΩ


12 VI0 = 4 kΩ

I0 = -3 mA based on direction


3 kΩ

9 kΩ 6 kΩ 12 kΩ

12 V

12 kΩ

4 kΩ

I0

Fig. (a)

3 kΩ


9 kΩ 6 kΩ 12 kΩ

12 V

4 kΩ 12 kΩ

I0

Fig. (b)


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


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 )


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


V3k = 6 × 100



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


9 kΩ 6 kΩ 12 kΩ

+ V0 -12 V

12 kΩ

4 kΩ


Fig. (a)

9 kΩ 6 kΩ 12 kΩ

12 V

4 kΩ 12 kΩ


12 kΩ × 4 kΩ= 12 kΩ + 4 kΩ

48 k × k


= 16 k

= 3 kΩ

A

9 kΩ 6 kΩ 12 kΩ

12 V C

3 kΩ

B Fig. (c)


3 kΩVCB = × VAB

3 kΩ + 9 kΩ


Here VAB = 12 V

3 kΩVCB = × 12 V 12 kΩ

VCB = 3 V = V0


4 kΩ 6 kΩ 12 kΩ

12 mA

6 kΩ

I0

Fig. (a)


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:


+6 V

V0 4 kΩ 4 kΩ 12 kΩ

- 4 kΩ

3 kΩ

6 kΩ

12 V

Fig. (a)

6 V 6 V


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


= 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

+


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


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


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)



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


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



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)



3 ΩV3Ω = × VBA

3 Ω + 6 Ω

Here VBA = 24 V

3 ΩV3Ω = × 24 V 9 Ω

V3Ω = 8 V = VCD


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


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


12 Ω × 12 Ω= 12 Ω + 12 Ω

144 Ω × Ω


= 24 Ω

= 6 Ω


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


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


3 Ω × 3 Ω= 3 Ω + 3 Ω


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


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:


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


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


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


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 + -


2 kΩ I1 I0 9 kΩ +

3 kΩ V0 = 6 V 4 kΩ

IS

-

B C Fig. (a)


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


VAB

I1 = 2 kΩ + 4 kΩ

24 VI1 = 6 kΩ

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 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


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


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)


V1

I10k = 10 kΩ

5 VI10k = 10 kΩ


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


VAB

I1 = 4 kΩ

6 VI1 = 4 kΩ

I1 = 1.5 mA


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


VS


V1

+ -

2 kΩ 4 kΩ

6 kΩ 4 kΩ 4 kΩ

VS


4 kΩ × 4 kΩ= 4 kΩ + 4 kΩ

16 k × k= 8 k

= 2 kΩ


V1

A + -

I1 2 kΩ 4 kΩ I2 +I3

6 kΩ 2 kΩ

VS

-

BFig. (c)


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


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)


V0


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


12 VI1 = 12 Ω

I1 = 1 A


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


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


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


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


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



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)


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


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


IS = 2 mA + 1 mA

IS = 3 mA


I0

4 kΩ 6 kΩ

3 kΩ 4 kΩ


12 kΩ12 mA

Fig. (a)

I0

4 kΩ 6 kΩ 4 kΩ

3 kΩ 12 kΩ 12 mA


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

I0

4 kΩ

3 kΩ 2 kΩ 12 mA


3 kΩ I0 = × 12 mA 4 kΩ + 2 kΩ + 3 kΩ

3 kΩ I0 = × 12 mA 9 kΩ


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


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


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


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


0 = 46 + VA

VA = -46 V

Basic Engineering Circuit Analysis

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