Engineering Board Exam Reviewer · RL Circuits · hard
In the series-parallel RL circuit shown, a 240 V (rms) source with =500 rad/s drives R1=25, , R2=120, , and L=400 mH. Resistor R2 is in parallel with the inductor. Find the total impedance of the circuit in polar form.
- 125 -25.1°,
- 125 25.1°,
- 247 54.1°,
- 128 0°,
Answer
B. 125 25.1°,
Solution
XL= L=(500)(400 mH)=200, . The parallel branch is Zp=R2(+jXL)/R2+jXL=88.2+j52.9, and ZT=R1+Zp=113+j52.9, =125 25.1°, .
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