Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 6)
6.
A 1.2 k
resistor is in series with a 15 mH coil across a 10 kHz ac source. The magnitude of the total impedance is

Discussion:
11 comments Page 2 of 2.
Siddharth said:
7 years ago
R = 1.2k ohm = 1.2 * 1000 = 1200 ohm
L = 15mH = 15 * 10^-3 = 0.015H
F = 10kHz = 10 * 1000 = 10,000 Hz
Impedance Z = (?)
XL = 2 * fL
XL = 2 * 3.14 * 10,000 * 0.015,
XL = 942..
Impedance Z = √ (R^2 + XL^2).
Z = √ (1200^2 + 942^2),
Z = 1525.86.
Ans is Z = 1526.
L = 15mH = 15 * 10^-3 = 0.015H
F = 10kHz = 10 * 1000 = 10,000 Hz
Impedance Z = (?)
XL = 2 * fL
XL = 2 * 3.14 * 10,000 * 0.015,
XL = 942..
Impedance Z = √ (R^2 + XL^2).
Z = √ (1200^2 + 942^2),
Z = 1525.86.
Ans is Z = 1526.
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