Engineering Board Exam Reviewer · RLC Circuits and Resonance · medium
In the series RLC circuit shown, a 29 V (rms) source with =500 rad/s is connected to R=42, , L=100 mH, and C=200, F. Find the voltage across the inductor.
- 5 V
- 25 V
- 20 V
- 14.2 V
Answer
B. 25 V
Solution
XL= L=(500)(100 mH)=50, and XC=1/ C= 1(500)(200, F)=10, . The net reactance is X=XL-XC=40, (inductive). |Z|=sqrt(R2+(XL-XC)2)=sqrt((42, )2+(40, )2)=58, . I=V/|Z|= 29 V58, =0.5 A. VL=IXL=(0.5 A)(50, )=25 V.
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