For the circuit shown in fig, current in inductance is 0.8A while that in capacitance is 0.6 A. What is the current drawn from the source?

Options

A. `0.1A`
B. `0.3A`
C. `0.6A`
D. `0.2A`

Answer

D

Explanation

if an ac source `E=(E_0) sin omega t ` is applied across an inductance and capacitance in parallel, the current in inductance will lag the applied voltage while that across the capacitor will lead, and so,
`I_(L)=(E_0)/(X_L) sin (omega t -(pi)/(2))=-0.8 sqrt(2) cos omega t`
`I_(C)=(V)/(X_C) sin (omega t +(pi)/(2))=-0.6 sqrt(2) cos omega t`
So the current drawn from the source
`I=I_(L)+I_(C ) = -0.2 sqrt(2) cos omega t`
so the current drawn from the source
`I=(I_L)+(I_C)=-0.2 sqrt(2) cos omega t `
`I_(rms)=(I_0)/(sqrt(2)) = (0.2 sqrt(2))/(sqrt(2)) =0.2 A` .