The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is

Question

The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is (A) 1 henry (B) 4 henry (C) 10-3 henry (D) 4 ×10-3 henry

Tardigrade Admin · Accepted Answer

e=-L (d i/d t) But e=4 V and (d i/d t)=(0-1/10-3)=-1/ 10-3 ∴ (-1/10-3)(-L) =⇒ L=4 × 10-3 henry

Q. The back emf induced in a coil, when current changes from $1$ ampere to zero in one milli-second, is $4$ volts, the self inductance of the coil is

1 henry

4 henry

$1 0^{- 3}$ henry

$4 \times 1 0^{- 3}$ henry

$e = - L \frac{d i}{d t}$
But $e = 4 V$ and $\frac{d i}{d t} = \frac{0 - 1}{1 0 ^{- 3}} = - 1/1 0^{- 3}$
$∴ \frac{- 1}{1 0 ^{- 3}} (- L)$
$=\Rightarrow L = 4 \times 1 0^{- 3}$ henry

Q. The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is

1 henry

4 henry

10−3 henry

4×10−3 henry

e=−Ldtdi​ But e=4V and dtdi​=10−30−1​=−1/10−3 ∴10−3−1​(−L) =⇒L=4×10−3 henry

Q. The back emf induced in a coil, when current changes from $1$ ampere to zero in one milli-second, is $4$ volts, the self inductance of the coil is

$1 0^{- 3}$ henry

$4 \times 1 0^{- 3}$ henry

$e = - L \frac{d i}{d t}$
But $e = 4 V$ and $\frac{d i}{d t} = \frac{0 - 1}{1 0 ^{- 3}} = - 1/1 0^{- 3}$
$∴ \frac{- 1}{1 0 ^{- 3}} (- L)$
$=\Rightarrow L = 4 \times 1 0^{- 3}$ henry