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A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source of 200 volt, 100 Hz is connected across the coil, the current in the coil will be
A
$\frac{20}{\sqrt{13}}$ A
B
2.0 A
C
4.0 A
D
8.0 A
Detailed Solution
Inductive reactance $X_L = 2\pi fL$ is proportional to frequency.
At 50 Hz, $X_L = 20\ \Omega$; at 100 Hz, $X_L' = 20\times\frac{100}{50} = 40\ \Omega$
The resistance does not depend on frequency: R = 30 $\Omega$.
Impedance at 100 Hz: $Z = \sqrt{R^2 + X_L'^2} = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50\ \Omega$
$I = \frac{V}{Z} = \frac{200}{50}$
I = 4.0 A
At 50 Hz, $X_L = 20\ \Omega$; at 100 Hz, $X_L' = 20\times\frac{100}{50} = 40\ \Omega$
The resistance does not depend on frequency: R = 30 $\Omega$.
Impedance at 100 Hz: $Z = \sqrt{R^2 + X_L'^2} = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50\ \Omega$
$I = \frac{V}{Z} = \frac{200}{50}$
I = 4.0 A