Question

If a circuit made up of a resistance $1\Omega$ and inductance 0.01H, and alternating emf 200V at 50Hz is connected, then the phase difference between the current and the emf in the circuit is:

• A. $ta{n}^{-1}\left(\pi \right)$
• B. $ta{n}^{-1}\left(\frac{\pi }{2}\right)$
• C. $ta{n}^{-1}\left(\frac{\pi }{4}\right)$
• D. $ta{n}^{-1}\left(\frac{\pi }{3}\right)$

Answer 1

The correct option is A $ta{n}^{-1}\left(\pi \right)$

The angle by which voltage leads current is given by
$tan\varphi =\frac{X_L}{R}R=1\Omega ,f=50Hz,X_L=\omega L,L={10}^{-2}H=tan\varphi =\frac{\omega L}{R}=\frac{2\pi fL}{R}=tan\varphi =\frac{2\varphi \times 50\times {10}^2}{1}=\pi =\varphi =ta{n}^{-1}\left(\pi \right)$