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(A) The induced e.m.f. $(e)$ in the secondary coil is given by the formula: $e = M \cdot \frac{dI}{dt}$, where $M$ is the coefficient of mutual induct

Vedclass · Accepted Answer

$3.75 	imes 10^{-3} 	ext{ s}$

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(A) The induced e.m.f. $(e)$ in the secondary coil is given by the formula: $e = M \cdot \frac{dI}{dt}$, where $M$ is the coefficient of mutual induction, $dI$ is the change in current, and $dt$ is the time interval.Given values are: $M = 3 	ext{ H}$, $e = 4 	ext{ kV} = 4000 	ext{ V}$, and $dI = I_1 - I_2 = 7 	ext{ A} - 2 	ext{ A} = 5 	ext{ A}$.Substituting these values into the formula: $4000 = 3 \cdot \frac{5}{dt}$.Rearranging for $dt$: $dt = \frac{3 \cdot 5}{4000} = \frac{15}{4000} 	ext{ s}$.$dt = 0.00375 	ext{ s} = 3.75 	imes 10^{-3} 	ext{ s}$.

The coefficient of mutual induction is $3 \text{ H}$ and induced e.m.f. across secondary is $4 \text{ kV}$. Current in primary is reduced from $7 \text{ A}$ to $2 \text{ A}$. The time required for the change of current is

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