SATHEE: alternating-currents Question 25

Alternating Currents

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alternating-currents Question 25

Question: Q. 14. The figure shows a series $L C R$ circuit with $L =$ $5.0 H, C = 80 μ F, R = 40 Ω$ connected to a variable frequency of $240 V$ source. Calculate

(i) The angular frequency of the source which drives the circuit at resonance.

(ii) The current at the resonating frequency.

(iii) The rms potential drop across the capacitor at resonance.

A [Delhi I, II, III 2012]

Show Answer

Solution:

Ans. (i)

(ii) Current at resonance

$I_{r m s} = \frac{V_{r m s}}{R} = \frac{240}{40} A = 6 A 1 / 2 + 1 / 2$

(iii) $V_{r m s}$ across a capacitor

$\begin{aligned} V_{r m s} & = I_{r m s} X_{C} & = 6 \times \frac{1}{50 \times 80 \times 10^{- 6}} V & = \frac{6 \times 10^{6}}{4 \times 10^{3}} V = \frac{6000}{4} V = 1500 V 1 / 2 \end{aligned}$

[CBSE Marking Scheme, 2012]

Alternating Currents

alternating-currents Question 25

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Alternating Currents

alternating-currents Question 25

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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