Engineering Board Exam Reviewer · RL Circuits · hard
In the series-parallel RL circuit shown, a 240 V (rms) source with =1000 rad/s drives R1=10, , R2=100, , and L=50 mH. Resistor R2 is in parallel with the inductor. Find the magnitude of the current drawn from the source.
- 1.99 A
- 4.39 A
- 4.8 A
- 24 A
Answer
C. 4.8 A
Solution
XL= L=(1000)(50 mH)=50, . The parallel branch is Zp=R2(+jXL)/R2+jXL=20+j40, and ZT=R1+Zp=30+j40, =50 53.1°, . I=V/|ZT|= 240 V50, =4.8 A.
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