MCQ
An alternating $e.m.f$. of angular frequency $\omega $ is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency
- A$\frac{\omega }{4}$
- B$\frac{\omega }{2}$
- C$\omega $
- ✓$2\omega $
✓
Answer
Correct option: D.
$2\omega $
d
(d)The instantaneous values of emf and current in inductive circuit are given by $E = {E_0}\sin \omega t$and $i = {i_0}\sin \left( {\omega t - \frac{\pi }{2}} \right)$respectively.
So, ${P_{inst}} = Ei = {E_0}\sin \omega t \times {i_0}\sin \left( {\omega t - \frac{\pi }{2}} \right)$
$ = {E_0}{i_0}\sin \omega t\left( {\sin \omega t\cos \frac{\pi }{2} - \cos \omega t\sin \frac{\pi }{2}} \right)$
$ = {E_0}{i_0}\sin \omega t\;\cos \omega t$
$ = \frac{1}{2}{E_0}{i_0}\sin 2\omega t$ $(\sin 2\omega t = 2\sin \omega t\;\cos \omega t)$
Hence, angular frequency of instantaneous power is $2\omega $.
(d)The instantaneous values of emf and current in inductive circuit are given by $E = {E_0}\sin \omega t$and $i = {i_0}\sin \left( {\omega t - \frac{\pi }{2}} \right)$respectively.
So, ${P_{inst}} = Ei = {E_0}\sin \omega t \times {i_0}\sin \left( {\omega t - \frac{\pi }{2}} \right)$
$ = {E_0}{i_0}\sin \omega t\left( {\sin \omega t\cos \frac{\pi }{2} - \cos \omega t\sin \frac{\pi }{2}} \right)$
$ = {E_0}{i_0}\sin \omega t\;\cos \omega t$
$ = \frac{1}{2}{E_0}{i_0}\sin 2\omega t$ $(\sin 2\omega t = 2\sin \omega t\;\cos \omega t)$
Hence, angular frequency of instantaneous power is $2\omega $.
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