Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
6th Edition
ISBN: 9780078028229
Author: Charles K Alexander, Matthew Sadiku
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 2, Problem 23P

In the circuit shown in Fig. 2.87, determine Vx and the power absorbed by the 60-Ω resistor.

Figure 2.87

For Prob. 2.23.

Chapter 2, Problem 23P, In the circuit shown in Fig. 2.87, determine Vx and the power absorbed by the 60- resistor. Figure

Expert Solution & Answer
Check Mark
To determine

Calculate the value of Vx and power absorbed by 60Ω resistor in Figure 2.87.

Answer to Problem 23P

The value of Vx is 100V_ and the power absorbed by 60Ω resistor is 960W_.

Explanation of Solution

Formula used:

Consider the expression for N resistors connected in parallel.

1Req=1R1+1R2+1R3++1RN

Here,

R1,R2,R3RN are resistors.

Consider the expression for N resistors connected in series.

Req=R1+R2+R3++RN

Calculation:

Refer to Figure 2.87 in the textbook for prob. 2.23.

In Figure 2.87, 40Ωand60Ω are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.

1Req1=140Ω+160Ω=3+2120Ω=5120Ω=124Ω

Simplify the equation as follows.

Req1=1(124Ω)=24Ω

As Req1 and 6Ω are connected in series, therefore the equivalent resistance for the series connected resistance are calculated as follows.

Req2=Req1+6Ω=24Ω+6Ω=30Ω

In Figure 2.87, 15Ωand30Ω are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.

1Req3=115Ω+130Ω=1+230Ω=330Ω=110Ω

Simplify the equation as follows.

Req3=1(110Ω)=10Ω

As Req3 and 20Ω are connected in series, therefore the equivalent resistance for the series connected resistance are calculated as follows.

Req4=Req3+20Ω=10Ω+20Ω=30Ω

As Req2andReq4 are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.

1Req5=1Req2+1Req4=130Ω+130Ω=1+130Ω=230Ω

Simplify the equation as follows.

Req5=1(115Ω)=15Ω

Redraw Figure 2.87 as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 2, Problem 23P

Apply KCL at node V1 of Figure 1.

60A+V110Ω+V1(5+Req5)=060A+V110Ω+V1(5+15Ω)=0 {Req5=15Ω}60A+V110Ω+V120Ω=0V1[110Ω+120Ω]=60A

Simplify the equation as follows.

V1(0.1+0.05)=60AV1(0.15)=60AV1=60A0.15V1=400V

Consider current through 5Ω resistor as i5, from Figure 1 the expression for i5 can be written as follows.

i5=V15Ω+Req5

Substitute 15Ω for Req5 and 400V for V1 as follows.

i5=400V5Ω+15Ω=400V20Ω=20A

From Figure 1, consider the expression for vx.

vx=(5Ω)i5

Substitute 20A for i5 as follows.

vx=(5Ω)(20A)=100V

From Figure 1, the voltage across VReq5=V1vx.

Substitute 100V for vx and 400V for V1 as follows.

VReq5=(400V)(100V)=400V+100V=300V

From voltage division rule, the voltage across 60Ω resistor is calculated as follows.

V60Ω=VReq5(Req16Ω+Req1)

Substitute 300V for VReq5 and 24Ω for Req1 as follows.

V60Ω=(300V)(24Ω6Ω+24Ω)=(300V)(24Ω30Ω)=240V

Write the expression to find the power absorbed by 60Ω resistor.

p60Ω=(V60Ω)260Ω

Substitute 240V for V60Ω to obtain the power absorbed by 60Ω resistor.

p60Ω=(240V)260Ω=57,60060W=960W

Conclusion:

Thus, the value of Vx is 100V_ and the power absorbed by 60Ω resistor is 960W_.

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Chapter 2 Solutions

Fundamentals of Electric Circuits

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