Formula used:
Consider the expression for N resistors connected in parallel.
1Req=1R1+1R2+1R3+⋯+1RN
Here,
R1, R2, R3⋯RN 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, 40 Ω and 60 Ω are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.
1Req1=140 Ω+160 Ω=3+2120 Ω=5120 Ω=124 Ω
Simplify the equation as follows.
Req1=1(124 Ω)=24 Ω
As Req1 and 6 Ω are connected in series, therefore the equivalent resistance for the series connected resistance are calculated as follows.
Req2=Req1+6 Ω=24 Ω+6 Ω=30 Ω
In Figure 2.87, 15 Ω and 30 Ω are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.
1Req3=115 Ω+130 Ω=1+230 Ω=330 Ω=110 Ω
Simplify the equation as follows.
Req3=1(110 Ω)=10 Ω
As Req3 and 20 Ω are connected in series, therefore the equivalent resistance for the series connected resistance are calculated as follows.
Req4=Req3+20 Ω=10 Ω+20 Ω=30 Ω
As Req2 and Req4 are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.
1Req5=1Req2+1Req4=130 Ω+130 Ω=1+130 Ω=230 Ω
Simplify the equation as follows.
Req5=1(115 Ω)=15 Ω
Redraw Figure 2.87 as shown in Figure 1.
Apply KCL at node V1 of Figure 1.
60 A+V110 Ω+V1(5+Req5)=060 A+V110 Ω+V1(5+15 Ω)=0 {∵Req5=15 Ω}60 A+V110 Ω+V120 Ω=0V1[110 Ω+120 Ω]=−60 A
Simplify the equation as follows.
V1(0.1+0.05) ℧=−60 AV1(0.15 ℧)=−60 AV1=−60 A0.15 ℧V1=−400 V
Consider current through 5 Ω resistor as i5, from Figure 1 the expression for i5 can be written as follows.
i5=V15 Ω+Req5
Substitute 15 Ω for Req5 and −400 V for V1 as follows.
i5=−400 V5 Ω+15 Ω=−400 V20 Ω=−20 A
From Figure 1, consider the expression for vx.
vx=(5 Ω)i5
Substitute −20 A for i5 as follows.
vx=(5 Ω)(−20 A)=−100 V
From Figure 1, the voltage across VReq5=V1−vx.
Substitute −100 V for vx and −400 V for V1 as follows.
VReq5=(−400 V)−(−100 V)=−400 V+100 V=−300 V
From voltage division rule, the voltage across 60 Ω resistor is calculated as follows.
V60 Ω=VReq5(Req16 Ω+Req1)
Substitute −300 V for VReq5 and 24 Ω for Req1 as follows.
V60 Ω=(−300 V)(24 Ω6 Ω+24 Ω)=(−300 V)(24 Ω30 Ω)=−240 V
Write the expression to find the power absorbed by 60 Ω resistor.
p60 Ω=(V60 Ω)260 Ω
Substitute −240 V for V60 Ω to obtain the power absorbed by 60 Ω resistor.
p60 Ω=(−240 V)260 Ω=57,60060 W=960 W
Conclusion:
Thus, the value of Vx is −100 V_ and the power absorbed by 60 Ω resistor is 960 W_.