Physics and Electrical Engineering Concepts

If the current changes from 5A to nill in five seconds in a coil of 500 volts, then the self-inductance is _____ Henry?

Answer: 500

Overview

To find the self-inductance of a coil, we can use the formula: L = (ΔΦ / ΔI), where L is the self-inductance in Henry (H), ΔΦ is the change in magnetic flux in Weber (Wb), and ΔI is the change in current in Ampere (A).
In this case, the current changes from 5A to 0A (nil) in 5 seconds, and the voltage is 500 volts.
Using the formula, we can calculate the self-inductance as L = (500 V) / (5 A / 5 s), which simplifies to L = 500 H.
Therefore, the self-inductance of the coil is 500 Henry.
This calculation is essential in understanding the behavior of electrical circuits and the properties of coils.
Self-inductance is a critical parameter in designing and analyzing electrical systems, and it plays a significant role in determining the performance of coils and other electromagnetic devices.

Explanation

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

L = (ΔΦ / ΔI)

Where:

L is the self-inductance in Henry (H),

ΔΦ is the change in magnetic flux in Weber (Wb),

and ΔI is the change in current in Ampere (A).

In this case, the current changes from 5A to 0A (nil) in 5 seconds. (ΔI) is therefore 5A - 0A = 5A.

To find the change in (ΔΦ), we can use Faraday's law of electromagnetic induction

ΔΦ = L * ΔI

Since the current changes from 5A to 0A in 5 seconds, the rate of  (ΔI) is 5A / 5s = 1A/s.

Substituting the values into the equation:

ΔΦ = L * ΔI

500 V = L * 1 A/s

Since the (V) is equal to  (L) and  (ΔI), we can rewrite the equation as:

L = 500 V / 1 A/s

Simplifying the equation gives us:

L = 500 H

Therefore, the self-inductance of the coil is 500 Henry (H).

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