Question

The instantaneous current and voltage in an A.C. circuit is given by $I=\sin \omega t$ and $E=150\cos \left(\omega t+\frac{\pi }{3}\right)$ respectively. What will be the phase difference between voltage and current in the circuit?

Answer 1

Step 1: Given data

Current in AC circuit: $I=\sin \omega t$

Voltage in AC circuit: $E=150\cos \left(\omega t+\frac{\pi }{3}\right)$

Step 2: Formula used

The equation of current given in terms of sine as $I=I_0\sin \left(\omega t-\varphi \right)$, where $I_0$ is amplitude of current, $\omega$ is angular frequency, $t$ is time variable and $\varphi$ is phase difference.

The equation of voltage given in terms of cosine as $E=E_0\sin \left(\omega t-\varphi \right)$, where $E_0$ is the amplitude of voltage, $\omega$ is angular frequency, $t$ is time variable and $\varphi$ is phase difference.

To convert a function in sine into cosine, the formula used is $\sin \theta =\cos \left(\frac{\pi }{2}-\theta \right)$.

Step 3: Calculate the phase difference

Convert the voltage into cosine terms using the formula.

$\begin{array}{rcl}E& =& 150\cos \left(\omega t+\frac{\pi }{3}\right)\\ E& =& 150\cos \left(\frac{\pi }{2}-\left(\omega t+\frac{\pi }{3}\right)-\frac{\pi }{2}\right)\\ E& =& 150\cos \left(\frac{\pi }{2}-\left(\omega t+\frac{\pi }{3}+\frac{\pi }{2}\right)\right)\\ & =& 150\sin \left(\omega t+\frac{\pi }{3}+\frac{\pi }{2}\right)\end{array}$

Now on comparing the waves $I=\sin \omega t$ and $\begin{array}{rcl}E& =& 150\sin \left(\omega t+\frac{\pi }{3}+\frac{\pi }{2}\right)\end{array}$, the phase difference is

$\begin{array}{rcl}\left(\frac{\pi }{3}+\frac{\pi }{2}+\omega t\right)-\omega t& =& \frac{2\pi +3\pi }{6}\\ & =& \frac{5\pi }{6}\end{array}$

Hence, the phase difference between voltage and current in the circuit $\frac{5\pi }{6}$.