Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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Chapter 9, Problem 1P

(a) Using the superposition theorem, determine the current through the 12 Ω resistor of Fig. 9.125.

b. Convert both voltage sources to current sources and recalculate the current to the 12 Ω resistor.

c. How do the results of parts (a) and (b) compare?

Chapter 9, Problem 1P, (a) Using the superposition theorem, determine the current through the 12  resistor of Fig. 9.125.

Fig. 9.125

Expert Solution
Check Mark
To determine

(a)

The current through 12Ω resistor using superposition theorem for the given network.

Answer to Problem 1P

The current through 12Ω resistor using superposition theorem for the given network is 0.1A in upward direction.

Explanation of Solution

Given:

The resistors values are R1=4Ω,R2=2Ω and R3=12Ω.

The voltage sources are E1=16V and E2=-10V.

Concept Used:

Elements in the series have the same current.

Elements in the parallels have the same voltage.

Superposition theorem states that in any linear bilateral network having more than one source response in any one of the branches is equal to algebraic sum of the responses caused by individual source while rest of the sources are replaced by their internal impedances.

Short circuit the voltage source and open circuit the current source.

If the resistors are in series, then the value of the equivalent resistance for N series resistors is

Req=R1+R2+.........+RN.

If the resistors are in parallel, then the value of the equivalent resistance for N series resistors is

1Req=1R1+1R2+.........+1RN.

The special case for only two parallel resistors is

Req=R1R2R1+R2.

Current: I=VR

Where V is the voltage.           R is the resistance..

Current division is used to express the current across one of several parallel resistors in terms of current across the combination:

If there are two resistors in parallel then the current across second resistor is

I2=I(R1R1 +R2)Where I is the source current.            I2 is the current across resistor R2.

Calculation:

Firstly, short circuit the voltage source

Total resitanceRT=R1+R2||R3=4+2||12RT=12(2)12+2+4RT=5.71ΩTotal current=165.71=2.8AUsing current division rule,Current through R3 resistor is2.822+12=0.4A in downward direction          ....(1)                     Secondly,short circuit the E1 voltage source.Total resitanceRT=R1||R3+R2=4||12+2RT=12(4)12+4+2RT=5ΩTotal current=105=2AUsing current division rule,Current through R3 resistor is2416=0.5A in upward direction       ...(2)From (1) and (2)Total current through R3 resistor is0.5-0.4=0.1A in upward direction.

Conclusion:

Hence, the current through 12Ω resistor using superposition theorem for the given network is 0.1A in upward direction.

Expert Solution
Check Mark
To determine

(b)

The current through 12Ω resistor using superposition theorem for the given network after converting both the voltage source to current source.

Answer to Problem 1P

The current through 12Ω resistor using superposition theorem for the given network after converting both the voltage source to current source is 0.1A in upward direction.

Explanation of Solution

Given:

The resistors values are R1=4Ω,R2=2Ω and R3=12Ω.

The voltage sources are E1=16V and E2=-10V.

Concept Used:

Elements in the series have the same current.

Elements in the parallels have the same voltage.

Superposition theorem states that in any linear bilateral network having more than one source response in any one of the branches is equal to algebraic sum of the responses caused by individual source while rest of the sources are replaced by their internal impedances.

Short circuit the voltage source and open circuit the current source.

If the resistors are in series, then the value of the equivalent resistance for N series resistors is

Req=R1+R2+.........+RN.

If the resistors are in parallel, then the value of the equivalent resistance for N series resistors is

1Req=1R1+1R2+.........+1RN.

The special case for only two parallel resistors is

Req=R1R2R1+R2.

Current: I=VR

Where V is the voltage.           R is the resistance..

Current division is used to express the current across one of several parallel resistors in terms of current across the combination:

If there are two resistors in parallel then the current across second resistor is

I2=I(R1R1 +R2)Where I is the source current.            I2 is the current across resistor R2.

Calculation:

Converting voltage source to current source.I1=E1R1=164=4AI2=E2R2=102=5AFirstly open circuit the I2 current source.Using current division rule,Current through R3 resistor is41 121 12+12+14=0.4A in downward direction          ....(1)                     Secondly,open circuit the I1 current source.Total resitanceRp=12||2Rp=12(2)12+2Rp=1.71ΩUsing current division rule,Current through 1.71Ω resistor is544+1.71=3.5A Current through R3 resistor is3.51212+2=0.5A in upward direction          ....(2)From (1) and (2)Total current through R3 resistor is0.5-0.4=0.1A in upward direction.

Conclusion:

Hence, the current through 12Ω resistor using superposition theorem for the given network after converting both the voltage source to current source is 0.1A in upward direction.

Expert Solution
Check Mark
To determine

(c)

Compare the results of both the part (a) and (b).

Answer to Problem 1P

The results are similar in both the part (a) and (b).

Explanation of Solution

Given:

The resistors values are R1=4Ω,R2=2Ω and R3=12Ω.

The voltage sources are E1=16V and E2=-10V.

Concept Used:

Elements in the series have the same current.

Elements in the parallels have the same voltage.

Superposition theorem states that in any linear bilateral network having more than one source response in any one of the branches is equal to algebraic sum of the responses caused by individual source while rest of the sources are replaced by their internal impedances.

Short circuit the voltage source and open circuit the current source.

Calculation:

The results are similar in both the part (a) and (b).

Conclusion:

Hence, the results are similar in both the part (a) and (b).

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Chapter 9 Solutions

Introductory Circuit Analysis (13th Edition)

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