A varying current in a coil changes from 10A tozero in 0.5 sec. If the average e.m.f induced inthe coil is 220V, the self-inductance of the coil is [1995]a)5 Hb)6 Hc)11 Hd)12 HCorrect answer is option 'C'. Can you explain this answer?

Question

A varying current in a coil changes from 10A tozero in 0.5 sec. If the average e.m.f induced inthe coil is 220V, the self-inductance of the coil is      [1995]a)5 Hb)6 Hc)11 Hd)12 HCorrect answer is option 'C'. Can you explain this answer?

Answer

To find the self-inductance of the coil, we can use the formula:

emf = -L * (change in current / change in time)

Given:
emf = 220V
change in current = 10A - 0A = 10A
change in time = 0.5s

Let's substitute the given values into the formula and solve for L:

220V = -L * (10A / 0.5s)

Simplifying the equation:

220V = -L * 20A/s

Dividing both sides by -20A/s:

-11V/s = L

Therefore, the self-inductance of the coil is 11H (Henry).

Explanation:

- The average emf induced in a coil is given by the formula: emf = -L * (change in current / change in time).
- In this case, the change in current is from 10A to 0A, and the change in time is 0.5s.
- Substituting these values into the formula, we get the equation: 220V = -L * (10A / 0.5s).
- Simplifying the equation, we find that the self-inductance of the coil is -11V/s.
- The negative sign indicates that the induced emf opposes the change in current.
- The unit of self-inductance is Henry (H).

Therefore, the self-inductance of the coil is 11H (Henry).

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