Question:

An inductor is connected to an ac source of frequency 50 Hz. The frequency of the instantaneous power developed in the circuit is

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In a purely inductive circuit connected to an AC source, the instantaneous power is the product of the instantaneous voltage and current. The voltage across an inductor leads the current by \( 90^\circ \). The frequency of the power waveform is twice the frequency of the voltage and current waveforms.
Updated On: May 12, 2025
  • \( 25 \) Hz
  • \( 50 \) Hz
  • \( 100 \) Hz
  • \( 200 \) Hz
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The Correct Option is C

Solution and Explanation

When an inductor is connected to an AC source, the voltage across the inductor leads the current through it by \( 90^\circ \) (\( \pi/2 \) radians).
Let the instantaneous current through the inductor be \( i(t) = I_m \sin(\omega t) \), where \( I_m \) is the peak current and \( \omega \) is the angular frequency of the AC source.
The frequency \( f \) is related to the angular frequency by \( \omega = 2\pi f \).
Given \( f = 50 \) Hz, so \( \omega = 100\pi \) rad/s.
The instantaneous voltage across the inductor is \( v(t) = L \frac{di}{dt} = L \frac{d}{dt} (I_m \sin(\omega t)) = L I_m \omega \cos(\omega t) \).
The instantaneous power \( p(t) \) developed in the circuit is the product of the instantaneous voltage and the instantaneous current: $$ p(t) = v(t) i(t) = (L I_m \omega \cos(\omega t)) (I_m \sin(\omega t)) $$ $$ p(t) = L I_m^2 \omega \sin(\omega t) \cos(\omega t) $$ Using the trigonometric identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \), we can write \( \sin(\omega t) \cos(\omega t) = \frac{1}{2} \sin(2\omega t) \).
$$ p(t) = \frac{1}{2} L I_m^2 \omega \sin(2\omega t) $$ The angular frequency of the instantaneous power is \( 2\omega \).
Since \( \omega = 2\pi f \), the frequency of the instantaneous power \( f_p \) is: $$ f_p = \frac{2\omega}{2\pi} = \frac{2(2\pi f)}{2\pi} = 2f $$ Given that the frequency of the AC source \( f = 50 \) Hz, the frequency of the instantaneous power is: $$ f_p = 2 \times 50 \text{ Hz} = 100 \text{ Hz} $$ The frequency of the instantaneous power developed in the circuit is 100 Hz.
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