Q. A $50\ volt$ potential difference is suddenly applied to a coil with $L = 5 \times 10^{- 3}henry$ and $R = 180\ ohm$. The rate of increase of current after $0.001\ second$ is
A) $27.3\ amp/sec$
B) $27.8\ amp/sec$
C) $2.73\ amp/sec$
D) None of the above
Explanation
(d) The rate of increase of current
$= \frac{di}{dt} = \frac{d}{dt}i_{0}\left( 1 - e^{- Rt/L} \right) = \frac{d}{dt}i_{0} - \frac{d}{dt}i_{0}e^{- Rt/L}$
$= 0 - i_{0}e^{- Rt/L}.\frac{d}{dt}\ \left( - \frac{Rt}{L} \right) = i_{0}\frac{R}{L}e^{- Rt/L}$
$= \frac{50}{180} \times \frac{180}{5 \times 10^{- 3}} \times e^{- (180 \times 0.001)/(5 \times 10^{- 3})} = 10^{4} \times e^{- 36}A/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.