Question

An alternating emf of angular frequency $\omega $ is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency:
(A) $\dfrac{\omega }{4}$
(B) $\dfrac{\omega }{2}$
(C) $\omega $
(D) $2\omega $

Answer 1

Hint: Here, we need to find the angular frequency of the instantaneous power developed in the circuit when an alternating emf is applied across an inductance. Whenever there is a need to find the power, you look for current and voltage as the power can be expressed in terms of these two quantities. Find the instantaneous current and voltage produced in the circuit and use the relation between them and power to find the angular frequency of the power.

Complete step by step answer:
When an inductor is connected with an alternating emf, current will flow through the circuit as it is closed. Now, in case of an inductor, the voltage lags the current or says that the current leads ahead of voltage by some angle known as the phase. If there is no resistance in the circuit, this phase angle is equal to $\dfrac{\pi }{2}$.
As we know, the alternating emf will be given as $V = {V_0}\sin \left( {\omega t} \right)$, here ${V_0}$ is the peak voltage. Since the current leads the voltage by an angle of $\dfrac{\pi }{2}$, we will have current given as $I = {I_0}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)$
Now, instantaneous power is equal to the product of instantaneous current and instantaneous voltage. Mathematically, $P = VI$.

Let us substitute the obtained values of current and voltage in the above equation of power, we will get $P = \left( {{V_0}\sin \left( {\omega t} \right)} \right)\left( {{I_0}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)} \right)$
$
\implies P = {V_0}{I_0}\sin \left( {\omega t} \right)\sin \left( {\omega t + \dfrac{\pi }{2}} \right) \\
\implies P = {V_0}{I_0}\sin \left( {\omega t} \right)\left( {\sin \left( {\omega t} \right)\cos \dfrac{\pi }{2} + \cos \left( {\omega t} \right)\sin \dfrac{\pi }{2}} \right) \\
 $
(we have used the property $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$)
As $\cos \dfrac{\pi }{2} = 0$ and $\sin \dfrac{\pi }{2} = 1$, we get,
$
 P = {V_0}{I_0}\sin \left( {\omega t} \right)\cos \left( {\omega t} \right) \\
\implies P = \dfrac{2}{2}{V_0}{I_0}\sin \left( {\omega t} \right)\cos \left( {\omega t} \right) \\
\implies P = \dfrac{{{V_0}{I_0}}}{2}\left( {2\sin \left( {\omega t} \right)\cos \left( {\omega t} \right)} \right) \\
 \therefore P = \dfrac{{{V_0}{I_0}}}{2}\sin \left( {2\omega t} \right) \\
 $
As you can see, the angular frequency of the instantaneous power is two times the angular frequency of the applied alternating emf.
Hence, the instantaneous power developed in the circuit has an angular frequency $2\omega $.

So, the correct answer is “Option D”.

Note:
Remember that the instantaneous power is always given as the product of instantaneous current and voltage. To solve these kinds of questions involving trigonometric ratios, keep track of the properties of these trigonometric ratios. Also keep in mind that the current in an inductive circuit leads the voltage by an angle of $\dfrac{\pi }{2}$.