The current in a self inductance L = 40 mH is to be increased uniformly from 1A to 11A in 4 ms. The e.m.f induced in inductor during this process is

Question

The current in a self inductance L = 40 mH is to be increased uniformly from 1A to 11A in 4 ms. The e.m.f induced in inductor during this process is (A) 0.4 V (B) 440 V (C) 40 V (D) 100 V

Tardigrade Admin · Accepted Answer

ε = L (di/dt) = ((40 × 10-3 )(11 - 1)/4 × 10-3) = 100 V

Q. The current in a self inductance L = 40 mH is to be increased uniformly from 1A to 11A in 4 ms. The e.m.f induced in inductor during this process is

0.4 V

440 V

40 V

100 V

$ε = L \frac{d i}{d t} = \frac{( 40 \times 1 0 ^{- 3} ) ( 11 - 1 )}{4 \times 1 0 ^{- 3}}$ = 100 V

Q. The current in a self inductance L = 40 mH is to be increased uniformly from 1A to 11A in 4 ms. The e.m.f induced in inductor during this process is

0.4 V

440 V

40 V

100 V

ε=Ldtdi​=4×10−3(40×10−3)(11−1)​ = 100 V

$ε = L \frac{d i}{d t} = \frac{( 40 \times 1 0 ^{- 3} ) ( 11 - 1 )}{4 \times 1 0 ^{- 3}}$ = 100 V