A coil of resistance 20 Ω, inductance 0.8 H is connected to a 400
| A coil of resistance 20 Ω, inductance 0.8 H is connected to a 400 V DC supply. The initial rate of change of current is
A. 250 A/sec
B. 500 A/sec
C. 300 A/sec
D. 600 A/sec
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
i(0-) = i(0+) = OA
Apply kVL
\(- 400 + 20i\left( {{0^ + }} \right) + 0.8\frac{{di\left( {{0^ + }} \right)}}{{dt}} = 0\)
\(- 400 + 0 + 0.8\frac{{di\left( {{0^ + }} \right)}}{{dt}} = 0\)
\(\frac{{di\left( {{0^ + }} \right)}}{{dt}} = \frac{{400}}{{0.8}} = 500\;{\rm{A}}/{\rm{sec}}\)
