A source of ac voltage $ v= v_0 \sin\omega t$, is connected across a pure inductor of inductance L. Derive the expressions for the instantaneous current in the circuit. Show that average power dissipated in the circuit is zero.
CBSE FOREIGN - SET 3 2017
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Given $V=V_0\sin\omega t$
$V=L\frac{di}{dt}\Rightarrow di=\frac{V}{L}dt$
$\therefore\text{ }di=\frac{\text{V}_0}{L}\sin\omega t\text{ }dt$
Integrating $i=-\frac{\text{V}_0}{wL}\cos\omega t$
$\therefore\text{}i=-\frac{\text{V}_0}{wL}\sin(\pi/2-\omega t)=I_0\sin(\pi/2-\omega t)$
where $I_0=\frac{V_0}{wL}$
Average power
$P_{av}=\int\limits_{0}^{T}vi\text{ }dt$
$=\frac{-V_0^2}{wL}\int_0^T\sin\omega t\cos\omega t\text{ }dt$
$=\frac{-V_0^2}{wL}\int_0^T\sin(2\text{ }\omega t)\text{ }dt$
= 0
$V=L\frac{di}{dt}\Rightarrow di=\frac{V}{L}dt$
$\therefore\text{ }di=\frac{\text{V}_0}{L}\sin\omega t\text{ }dt$
Integrating $i=-\frac{\text{V}_0}{wL}\cos\omega t$
$\therefore\text{}i=-\frac{\text{V}_0}{wL}\sin(\pi/2-\omega t)=I_0\sin(\pi/2-\omega t)$
where $I_0=\frac{V_0}{wL}$
Average power
$P_{av}=\int\limits_{0}^{T}vi\text{ }dt$
$=\frac{-V_0^2}{wL}\int_0^T\sin\omega t\cos\omega t\text{ }dt$
$=\frac{-V_0^2}{wL}\int_0^T\sin(2\text{ }\omega t)\text{ }dt$
= 0
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