In a coil of self inductance of $5$ henry, the rate of change of current is $2$ ampere per second, the e.m.f. induced in the coil is

5 V

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- 5 V

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- 10 V

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10 V

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Solution

The correct option is C $- 10 V$
e.m.f. $= - L \frac{d i}{d t}$
$= - 5 \times 2$
$= - 10 V$

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In a coil of self inductance of 5 henry, the rate of change of current is 2 ampere per second, the e.m.f. induced in the coil is

The correct option is C −10Ve.m.f.=−Ldidt=−5×2=−10V

In a coil of self inductance of $5$ henry, the rate of change of current is $2$ ampere per second, the e.m.f. induced in the coil is

The correct option is C $- 10 V$
e.m.f. $= - L \frac{d i}{d t}$
$= - 5 \times 2$
$= - 10 V$