Question

The current flowing through the resistor in a series LCR a.c. circuit, is $I=\epsilon /R.$
Now the inductor and capacitor are connected in parallel and joined in series with the resistor as shown in figure. The current in the circuit is now. (Symbols have their usual meaning)

• A. equal to I
• B. more than I
• C. less than I
• D. zero

Answer 1

The correct option is D zero

In figure (1) $Z=R$

$X_L=X_C$

If $I_1=\frac{V_0}{X_C}sin(\omega t+\pi /2)$

then $I_2=\frac{V_0}{X_L}sin(\omega t-\pi /2)$

and Total current in circuit two $I=I_1+I_2=0$