Question

In a coil, an increase in current from 5 A to 10 A in 100 ms induces an emf of 100 V. The self-inductance of the coil is:

• A. 2 H
• B. 10 H
• C. 20 H
• D. 2000 H

Answer 1

The Correct Option is A

The induced electromotive force (EMF) in a coil due to a changing current is given by Faraday's Law of Induction and is related to the self-inductance (L) by the following formula:

\( EMF = -L \frac{\Delta I}{\Delta t} \)

Where:

• \( EMF \) is the induced electromotive force (in Volts).
• \( L \) is the self-inductance of the coil (in Henries).
• \( \Delta I \) is the change in current (in Amperes).
• \( \Delta t \) is the change in time (in seconds).

 

We are given:

• \( EMF = 100 \, \text{V} \)
• \( \Delta I = 10 \, \text{A} - 5 \, \text{A} = 5 \, \text{A} \)
• \( \Delta t = 100 \, \text{ms} = 0.1 \, \text{s} \)

 

We need to find the self-inductance \( L \). Rearranging the formula to solve for \( L \):

\( L = -\frac{EMF}{\frac{\Delta I}{\Delta t}} = -\frac{EMF \cdot \Delta t}{\Delta I} \)

Substituting the given values:

\( L = -\frac{100 \, \text{V} \cdot 0.1 \, \text{s}}{5 \, \text{A}} = -\frac{10}{5} \, \text{H} = -2 \, \text{H} \)

Since self-inductance \( L \) is always a positive quantity, we take the absolute value:

\( L = |-2 \, \text{H}| = 2 \, \text{H} \)

Therefore, the self-inductance of the coil is 2 H.

Answer 2

\(\text{ The induced emf } \varepsilon \text{ in a coil is related to the self-inductance } L \text{ by the formula:}\\\)

\[\varepsilon = L \frac{dI}{dt}\]

\(\text{where } \frac{dI}{dt} \text{ is the rate of change of current. In this case, } \frac{dI}{dt} = \frac{10 - 5}{0.1} = 50 \, \text{A/s}.\\\)
\(\text{Substituting the values:}\\\)
\(100 = L \times 50 \quad \Rightarrow \quad L = \frac{100}{50} = 2 \, H\)
\(\text{Thus, the self-inductance is } 2 \, H.\)