Question

An $AC$ voltage $E=220\sin (\omega t-\frac{\pi }{6})$ is connected across a certain device. The current flowing through this device is $i=115\sin (\omega t+\frac{\pi }{3})$. The correct phasor diagram, of applied voltage and current flowing in the circuit, is,

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Given:
$E=220\sin (\omega t-\frac{\pi }{6})$

$i=115\sin (\omega t+\frac{\pi }{3})$

From the equations of current and voltage, we can say that the current is leading the voltage.

The phase difference between the current and voltage is,

$\varphi =\frac{\pi }{3}-\left(\frac{\pi }{6}\right)=\frac{\pi }{2}$

Also, the magnitude of voltage is more than the magnitude of current.

Hence, the phasor diagram can be drawn as follows,


Hence, $\left(C\right)$ is the correct answer.