SATHEE: alternating-currents Question 24

alternating-currents Question 24

Question: Q. 12. A voltage $V = V_{0} \sin ω t$ is applied to a series $L C R$ circuit. Derive the expression for the average power dissipate over a cycle.

Under what conditions is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit ?

U[O.D. I, II, III 2014]

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Solution:

Ans.

Applied voltage $= V_{0} \sin ω t$

$1 / 2$

Current in the circuit $= I_{0} \sin (ω t - ϕ)$

where, $ϕ$ is the phase lag of the current with respect to the voltage applied.

Hence instantaneous power dissipation

$\begin{aligned} = V_{0} \sin ω t \times I_{0} \sin (ω t - ϕ) & = \frac{V_{0} I_{0}}{2} [2 \sin ω t \cdot \sin (ω t - ϕ)] & = \frac{V_{0} I_{0}}{2} [\cos ϕ - \cos (2 ω t - ϕ)]^{1 / 2} \end{aligned}$

Therefore, average power for one completes cycle $=$ average of $[\frac{V_{0} I_{0}}{2} \cos ϕ - \cos (2 ω t - ϕ)$

The average of the second term over a complete cycle is zero.

Hence, average power dissipated over one complete cycle $= \frac{V_{0} I_{0}}{2} \cos ϕ \times 1 / 2$ Conditions :

(i) No power is dissipated when $R = 0 {(or ϕ = 90^{\circ})}^{1 / 2}$ [Note : Also accept if the student writes ‘This condition cannot be satisfied for a series LCR circuit’.]

(ii) Maximum power is dissipated when $X_{L} = X_{C} \cdot 1 / 2$ Or

$ω L = \frac{1}{ω C} (or ϕ = 0)$

[CBSE Marking Scheme 2014]

Alternating Currents

alternating-currents Question 24

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Alternating Currents

alternating-currents Question 24

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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