Question

plzz answer this with explaination..... i will mark as brainliest.... ​

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf Instantaneous\:Power(P)=\frac{-V^2sin2 \omega t}{2L\omega} }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• V = $\sf{V_osin\omega t}$

$\large\underline\pink{\sf To\:Find: }$

• Instantaneous Power (P) = ?

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If $\sf{\frac{dl}{dt}}$ is rate of change of current through inductor L.

Then ,

Induced emf is in the inductor of the same instant is :-

$\large\implies{\sf \left(-L\frac{dI}{dt}\right)}$

To maintain the flow of current we must apply voltage equal and opposite to the induced voltage .

therefore,

$\large\implies{\sf V=-\left(-L\frac{dI}{dt}\right)}$

Given :- V = $\sf{V_osin \omega t}$

$\sf{V_osin \omega t= L\frac{dI}{dt}→(1)}$

$\large\implies{\sf \frac{dI}{dt}=\frac{V_o}{L}sin \omega t }$

$\large\implies{\sf I = -\frac{V_o}{L}\frac{cos \omega t}{\omega} →(2)}$

$\large{\boxed{\sf Instantaneous\:Power(P)=VI}}$

$\large\implies{\sf P=V_osin \omega t \left(-\frac{Vcos\omega t}{L\omega}\right) }$

$\large\implies{\sf P=\frac{V^2sin2\omega t}{2L\omega} }$

Hence ,

$\red{\boxed{\sf Instantaneous\:Power(P)=\frac{-V^2sin2 \omega t}{2L\omega} }}$