Question 246: Electromagnetic Induction

Question

The coefficient of mutual induction is 2 H and induced e.m.f. across secondary is 2 kV Current in the primary is reduced from 6 A to 3 A . The time required for the change of current is

Accepted Answer

To determine the time required for the change in current, we can use the formula related to mutual induction:
$$ \varepsilon = -M \frac{\Delta I}{\Delta t} $$
where:
$\varepsilon$ is the induced electromotive force (e.m.f) across the secondary coil,
$M$ is the mutual inductance,
$\Delta I$ is the change in current in the primary coil,
$\Delta t$ is the time over which the change occurs.
Given:
$\varepsilon = 2 \, 	ext{kV} = 2000 \, 	ext{V}$
$M = 2 \, 	ext{H}$
The current change $\Delta I = 6 \, 	ext{A} - 3 \, 	ext{A} = 3 \, 	ext{A}$
Plug these values into the formula:
$$ 2000 = -2 \cdot \frac{3}{\Delta t} $$
Solving for $\Delta t$:
First, eliminate the negative sign (since we are considering magnitude):
$$ 2000 = 2 \cdot \frac{3}{\Delta t} $$
Multiply both sides by $\Delta t$:
$$ 2000 \Delta t = 6 $$
Now, solve for $\Delta t$:
$$ \Delta t = \frac{6}{2000} = 0.003 \, 	ext{s} = 3 	imes 10^{-3} \, 	ext{s} $$
Thus, the time required for the change of current is:
Option A
: $3 	imes 10^{-3} \, 	ext{s}$

Answer

$3 	imes 10^{-3} \mathrm{~s}$

Answer

$3 	imes 10^{-2} \mathrm{~s}$

Answer

$6 	imes 10^{-3} \mathrm{~s}$

Answer

$3 	imes 10^{-2} \mathrm{~s}$

Electromagnetic Induction

The coefficient of mutual induction is 2 H and induced e.m.f. across secondary is 2 kV Current in the primary is reduced from 6 A to 3 A . The time required for the change of current is

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