Question

In an induction coil, the coefficient of mutual inductance is 4 H. If a current of 5 A in the primary circuit is cut off in $\frac{1}{500}$s, the emf induced in the secondary circuit will be

• A. 30 KV
• B. 20 KV
• C. 10 KV
• D. 5 KV

Answer 1

The correct option is C 10 KV

$e=-M\frac{di}{dt}$
$e=-4\left[\frac{5}{\left(\frac{1}{500}\right)}\right]$
$=+4\times 5\times 500=20\times 500=20\times 500$
$=10\times {10}^3$
$=10kV$