Question

In the given circuit, peak voltage across the resistor and capacitor are 6V and 8V. At an instant when voltage across capacitor is 4V in magnitude,

• A. Voltage across resistance is $-3\sqrt{3}V$
• B. Voltage across resistance is $3\sqrt{3}V$
• C. Peak source voltage is $5\sqrt{3}$
• D. Current in circuit is ahead of source voltage

Answer 1

Answer: (A), (B), (D)

The correct options are
A Voltage across resistance is $-3\sqrt{3}V$
B Voltage across resistance is $3\sqrt{3}V$
D Current in circuit is ahead of source voltage

$V_R=V_{Rmax}\sin (\omega t+\theta )=6\sin (\omega t+\theta )$
$V_C=-V_{Cmax}\cos (\omega t+\theta )=-8\cos (\omega t+\theta )$
At an instant, voltage across the capacitor is $4V$
$\Rightarrow -8\cos (\omega t+{53}^{\circ })=4\cos (\omega t+53)=-\frac{1}{2}\sin (\omega t+53)=\pm \frac{\sqrt{3}}{2}\Rightarrow V_R=6\sin (\omega t+53)=\pm 6\times \frac{\sqrt{3}}{2}=\pm 3\sqrt{3}$
$V_{sourcemax}=\sqrt{V_{Rmax}^2+V_{Cmax}^2}$
= $10V$
In an RC circuit, current always leads voltage.