Q. In a L – R circuit, the value of L is $\left( \frac{0.4}{\pi} \right)$H and the value of R is $30\ \Omega$. If in the circuit, an alternating emf of 200 V at 50 cycle/s is connected, the impedance of the circuit and current will be
A) $11.4\ \Omega,\ 17.5\ A$
B) $30.7\ \Omega,\ 6.5\ A$
C) $40.4\ \Omega,\ 5\ A$
D) $50\ \Omega,\ 4\ A$
Explanation
(d) $Z^{2} = R^{2} + (2\pi fL)^{2}$
$= (30)^{2} + \left( 2\pi \times 50 \times \frac{0.4}{\pi} \right)^{2}$
$= (900 + 1600) = 2500$
or $Z = 50\ \Omega$
Also, $I = \frac{V}{Z} = \frac{200}{50} = 4\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.