SATHEE: alternating-currents Question 27

alternating-currents Question 27

Question: Q. 2. An $a c$ voltage $V = V_{0} \sin ω t$ is applied to a pure inductor $L$ . Obtain an expression for the current in the circuit. Prove that the average power supplied to an inductor over one complete cycle is zero.

A [SQP I 2017]

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Solution:

Ans. Expression for current 3

Proof 2

(i) If an alternating voltage $V = V_{0} \sin ω t$ is applied across pure inductor of inductance $L$ , then the magnitude of induced emf will be equal to applied voltage,

$E = L \frac{d I}{d t}$

For the circuit, magnitude of induced emf

$=$ applied voltage

$\begin{aligned} L \frac{d I}{d t} & = V_{0} \sin ω t d I & = \frac{V_{0}}{L} \sin ω t d t \end{aligned}$

Integrating both the sides, we get

$\begin{aligned} I = \frac{V_{0}}{L} \int \sin ω t d t = \frac{V_{0}}{L} (\frac{- \cos ω t}{ω}) & I = - \frac{V_{0}}{ω L} \cos ω t = - \frac{V_{0}}{ω L} \sin (\frac{π}{2} - ω t) & I = \frac{V_{0}}{X_{L}} \sin (ω t - \frac{π}{2}) if X_{L} = ω L & I = I_{0} \sin (ω t - \frac{π}{2}) \end{aligned}$

(ii) The average power supplied by the source over a complete cycle is

$P_{a v} = E_{r m s} I_{r m s} \cos ϕ$

When the circuit carries an ideal inductor, then the phase difference between the current and voltage is $\frac{π}{2}$

In case of pure inductive circuit,

$\begin{aligned} ϕ & = \frac{π}{2} But \cos \frac{π}{2} & = 0 \end{aligned}$

So power dissipated $= 0$

Hence, when an ac source is connected to an ideal inductor, the average power supplied by the source over a complete cycle is zero.

[CBSE Marking Scheme 2017]

Alternating Currents

alternating-currents Question 27

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Alternating Currents

alternating-currents Question 27

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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