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Chapter 16: Problem 19

An average EMF of \(20 \mathrm{~V}\) is induced in an inductor when the current in it changed from \(2.5 \mathrm{~A}\) in one direction to the same value in opposite direction in \(0.1 \mathrm{~s}\), the self-inductance of inductor is (A) \(0.4 \mathrm{H}\) (B) \(1 \mathrm{H}\) (C) \(2 \mathrm{H}\) (D) \(0.6 \mathrm{H}\)

Short Answer

Expert verified
The self-inductance of the inductor is 2 H, so the correct answer is (C) \(2 \mathrm{H}\).

Step by step solution

01

Understand the given data and variables

We are given the following: - Average EMF = 20 V - Change in current: from 2.5 A in one direction to 2.5 A in the opposite direction -> total change = 5 A - Time taken for the current change = 0.1 s We need to find the self-inductance \(L\) of the inductor.
02

Rearrange the formula for the induced EMF to solve for L

The formula for the induced EMF is: \[EMF = L\frac{\Delta I}{\Delta t}\] We need to solve for \(L\). To do this, let's rearrange the formula: \[L = \frac{EMF \times \Delta t}{\Delta I}\]
03

Substitute the given values into the formula

Now plug in the given values into the formula: \[L = \frac{20 \text{ V} \times 0.1 \text{ s}}{5 \text{ A}}\]
04

Calculate the value of L

Calculate the self-inductance: \[L = \frac{20 \text{ V} \times 0.1 \text{ s}}{5 \text{ A}} = \frac{2 \text{ V} \cdot \text{s}}{1 \text{ A}} = 2 \text{ H}\] The self-inductance of the inductor is 2 H, so the correct answer is (C) \(2 \mathrm{H}\).

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