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.