Question

In an AC circuit, voltage V=${V_0}\sin \omega t$ and inductor L is connected across the circuit. Then the instantaneous power will be
A). $\dfrac{{V_0^2}}{{2\omega L}}\sin \omega t$
B). $\dfrac{{ - V_0^2}}{{2\omega L}}\sin \omega t$
C). $- \dfrac{{V_0^2}}{{2\omega L}}\sin 2\omega t$
D). $\dfrac{{V_0^2}}{{\omega L}}\sin \omega t$

Answer 1

In devices where the current and voltage are not constant, like in the case of AC devices, the power dissipated can be calculated as a function of time as current and voltage changes with time. Instantaneous power is the power dissipated by the body at a given instant of time.

Complete step-by-step solution:
According to this question the voltage V=${V_0}\sin \omega t$…………………………. (Given)
Where inductor L is also given.
If $\dfrac{{di}}{{dt}}$ is the rate of current through inductor L at any instant then induced emf in the inductor of same instant is
$= \left( { - L\dfrac{{di}}{{dt}}} \right)$
Now as know that induced emf is equal to
$e = - \left( { - L\dfrac{{di}}{{dt}}} \right)$
Now there is no resistance, so
${({V_{Net}})_{\operatorname{int} }} = {V_0}\sin \omega t - L\dfrac{{di}}{{dt}} = 0$

This is because of no resistance presence of circuit in it
Now, $L\dfrac{{di}}{{dt}} = {V_0}\sin \omega t$
$\Rightarrow di = \dfrac{{{V_0}}}{L}\sin \omega t$
Now integrating both sides
$\Rightarrow \int {di} = \dfrac{{{V_0}}}{L}\int {\sin \omega t} dt$
$\begin{gathered} \Rightarrow i = \dfrac{{{V_0}}}{L}\left( { - \dfrac{{\cos \omega t}}{\omega }} \right) \\\ = - \dfrac{{{V_0}}}{{\omega L}}\cos \omega t \\\ \end{gathered}$
Therefore instantaneous power,
${P_{inst}} = {V_{inst}} \times {i_{inst}} = {V_0}\sin \omega t\left( { - \dfrac{{{V_0}}}{{\omega L}}\cos \omega t} \right)$
${P_{inst}} = - \dfrac{{V_0^2}}{{\omega L}}\sin \omega t \cdot \cos \omega t$
Now,
${P_{inst}} = - \dfrac{{V_0^2}}{{\omega L}}\dfrac{2}{2}\sin \omega t \cdot \cos \omega t$
since$(\sin 2\theta = 2\sin \theta \cdot \cos \theta )$therefore
${P_{inst}} = - \dfrac{{V_0^2}}{{2\omega L}}\sin 2\omega t$ …………………………………. (eq)
The given equation (eq) shows the instantaneous power.
Hence, option (C) is the correct answer.

Instantaneous power p (t) is the power at any instant of time. It is measured in watts. Power is the work done in a unit of time or we can also say that power is measured in terms of how quickly work can be done.

Note: A unit of power is equivalent to a unit of work divided by a unit of time. One horsepower Equivalent to approximately 750 Watt. Work and power are two different terms where the work is the product where the external force is applied on the object on the other side the power is measured on the amount of work done per unit of time.