Question

A segment of a circuit is shown in figure $V_R=5V,V_c=4sin2t.$ The voltage $V_L$ is given by

• A. 3 - 8 cos 2t
• B. 32 sin 2t
• C. 16 sin 2t
• D. 16 cos 2t

Answer 1

The correct option is B 32 sin 2t

By KCL,

$I_P+I_Q+I_C+I_L=0$

$2+1+I_C$ + $I_L$ = 0

But,

$I_C=C\times \frac{dv}{dt}$

$=1\times \frac{d}{dt}\left(4sin2t\right)=8cos2t$

$\therefore I_L=-(2+1+8cos2t)$

$=-3-8cos2t$

$\therefore V_L=L\frac{di}{dt}=2\times 2\times 8sin2t$

$=32sin2t$

Note: KCL is based on the law of conservation of charges.