RL Circuits - Engineering

Q1:

A 140  resistor is in parallel with an inductor having 60  inductive reactance. Both components are across a 12 V ac source. The magnitude of the total impedance is

A 5.51 

B 55.15 

C 90 

D 200 

ANS:B - 55.15 

Z = R * XL/√(R2 * XL2).
Z= 140 * 60/√(140 * 140 * 60 * 60).
Z= 55.15Ω.

Option B.