Question

For the circuit shown in the Figure, the current through the inductor is 0.9 A while the current through the condenser is 0.4 A

• A. current drawn from generator I = 1.13 A
• B. $\omega =\frac{1}{\left(1.5LC\right)}$
• C. I = 0.5 A
• D. I = 0.6 A

Answer 1

The correct option is C I = 0.5 A

The current in inductor and capacitor is always at an phase difference of ${180}^o$
for $V=V_{o}sin\omega t$
Capacitor, $i=i_{o}sin(\omega t-\frac{\pi }{2})$
Inductor, $i=i_{o}sin(\omega t+\frac{\pi }{2})$
So, the current from both branches will be $0.9+(-0.4)=0.5A$