Question

A 10 Ω, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 μs. The average e.m.f. induced in the coil is ____V

Answer 1

Correct Answer: 400

To find the average induced electromotive force (e.m.f.) in the coil, we use Faraday's law of electromagnetic induction, which states that the induced e.m.f. (𝓔) is given by the rate of change of magnetic flux. For a coil, the e.m.f. can be expressed in terms of the rate of change of current (di/dt) as follows:

𝓔 = -L(di/dt)

Where:
L = inductance of the coil = 20 mH = 20 × 10^-3 H,
di = change in current = initial current - final current.

When the switch is opened, the current changes from its initial value to zero:
di = 0 - I, where I is the initial current.

The time interval (dt) in which the current changes is 100 μs = 100 × 10^-6 s.

The average current (I) can be found using Ohm's Law:
I = V/R, where V = 20 V and R = 10 Ω.
Therefore, I = 20 V / 10 Ω = 2 A.

Now substituting the values into the e.m.f. equation:
𝓔 = -L(di/dt) = -20 × 10^-3 H × (0 - 2 A) / (100 × 10^-6 s)
= (20 × 10^-3 H) × (2 A) / (100 × 10^-6 s)
= 400 V.

Therefore, the average induced e.m.f. in the coil is 400 V, which fits within the given range of 400 to 400 V.