SATHEE: alternating-currents Question 17

alternating-currents Question 17

Question: Q. 4. The figure shows a series $L C R$ circuit connected to a variable frequency of $200 V$ source with $L = 50 mH, C = 80 μ F$ and $R = 40 Ω$ find.

(i) the source frequency which drives the circuit in resonance;

(ii) the quality factor $(Q)$ of the circuit.

A [O.D. Comptt. I, II, III 2014]

Show Answer

Solution:

Ans. (i) $ω_{0} = \frac{1}{\sqrt{L C}}$

$ω_{0} = \frac{1}{\sqrt{50 \times 10^{- 3} \times 80 \times 10^{- 6}}} = 500 rad / s$

(ii)

[CBSE Marking Scheme 2014]

Short Answer Type Questions-II

[AI Q. 1. A source of $a c$ voltage $V = V_{0} \sin ω t$ , is connected across a pure inductor of inductance $L$ . Derive the expressions for the instantaneous current in the circuit. Show that average power dissipated in the circuit is zero.

[Foreign I, II, III 2017]

Ans. Derivation of instantanteous current Derivation of average power dissipated

Given : $V = V_{0} \sin ω t$

$V = L \frac{d i}{d t} \Rightarrow d i = \frac{V}{L} d t$

$\begin{aligned} ∴ d i = \frac{V_{0}}{L} \sin ω t d t & Integrating, i = - \frac{V_{0}}{ω L} \cos ω t & ∴ i = - \frac{V_{0}}{ω L} \sin (\frac{π}{2} - ω t) = I_{0} \sin (\frac{π}{2} - ω t) \end{aligned}$

where,

$I_{0} = \frac{V_{0}}{ω L}$

Average power,

$\begin{aligned} (1) & P_{a v} & = \int_{0}^{T} V i d t & = \frac{- V_{0}^{2}}{ω L} \int_{0}^{T} \sin ω t \cos ω t d t & = \frac{- V_{0}^{2}}{2 ω L} \int_{0}^{T} \sin (2 ω t) d t & = 0 \end{aligned}$

[CBSE Marking Scheme 2017]

Alternating Currents

alternating-currents Question 17

Short Answer Type Questions-II

Ans. Derivation of instantanteous current Derivation of average power dissipated

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Alternating Currents

alternating-currents Question 17

Short Answer Type Questions-II

Ans. Derivation of instantanteous current Derivation of average power dissipated

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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