Question

In a series ac circuit, the instantaneous voltage V(t) and current I(t) are given by $V\left(t\right)=5[cos\omega t+\sqrt{3}sin\omega t]$
$I\left(t\right)=5\left[sin\right(\omega t+\frac{\pi }{4}\left)\right],$ then

• A. Circuit is in resonance
• B. Circuit is more inductive
• C. Circuit is more capacitive
• D. /Information is insufficient

Answer 1

The correct option is C

Circuit is more capacitive

$V\left(t\right)=\left(5\right)\left(2\right)\{\frac{1}{2}cos\omega t+\frac{\sqrt{3}}{2}sin\omega t\}$
$V\left(t\right)=10\left[sin\right(\omega t+{30}^{\circ }\left)\right]$
$I\left(t\right)=5\left[sin\right(\omega t+{45}^{\circ }\left)\right]$
Circuit is more capacitive as current leads voltage.