Question

When current is coil changes from 5 A to 2 A in 0.1 s, an average of 50 V is produced. The self-inductance of the coil is.

• A. 6 H
• B. 0.67 H
• C. 1.67 H
• D. 3 H

Answer 1

The correct option is C 1.67 H

Inudctor opposes change in curren, as the current is decreased, the inductor will convert magnetic energy into electrical by creating a potential difference. The voltage drop across inductor is given by

$V=L\frac{di}{dt}$

$-50=L\frac{2-5}{.1}$ here negative sign is because voltage is increasing instead of drop across inductor

$-50=L\frac{2-5}{.1}$

$L=1.67H$.