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RL Circuits - Engineering

Q1:

A 12 k resistor is in series with a 90 mH coil across an 8 kHz ac source. Current flow in the circuit, expressed in polar form, is I = 0.30° mA. The source voltage, expressed in polar form, is

A 3.8420.6° V

B 12.820.6° V

C 0.320.6° V

D 13.8469.4° V

ANS:A - 3.8420.6° V

Source voltage = current * impedence.
Impedence = R+jX ( X is reactance of the inductor i.e., X=2*3.14*f*L= 4521.6ohm).
Impedence = 12000+j4521.6.
in polar form Impedence = 12823.60<20.64.
Hence Voltage = 0.3*10^(-3)<0 * 12823.60<20.64 = 3.847<20.646.
A is correct.