Engineering Board Exam Reviewer · RL Circuits · medium
In the series RL circuit shown, a 170 V (rms) source with =2000 rad/s is connected to R=150, and L=40 mH. Find the real power dissipated by the circuit.
- 170 W
- 80 W
- 193 W
- 150 W
Answer
D. 150 W
Solution
XL= L=(2000)(40 mH)=80, . |Z|=sqrt(R2+XL2)=sqrt((150, )2+(80, )2)=170, . I=V/|Z|= 170 V170, =1 A. Only the resistor dissipates real power: P=I2R=(1 A)2(150, )=150 W.
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