In a LCR circuit capacitance is changed from \(\mathrm{C}\) to \(2 \mathrm{C}\). For the resonant frequency to remain unchanged, the inductance should be change from \(L\) to (a) \(4 \mathrm{~L}\) (b) \(2 \mathrm{~L}\) (c) \(L / 2\) (d) \(\mathrm{L} / 4\)

Short Answer

Expert verified
The inductance should change from L to L/2 for the resonant frequency to remain unchanged. The correct answer is (c) L/2.

Step by step solution

01

Formulate resonance frequency formulas for the initial and final LCR circuits

For a series LCR circuit, the resonance frequency (\(f_0\)) is given by the formula: \(f_0 = \dfrac{1}{2 \pi \sqrt{L C}}\) For the initial LCR circuit, we have capacitance C and inductance L: \(f_0 = \dfrac{1}{2 \pi \sqrt{L C}}\) For the final LCR circuit, the capacitance is doubled, and we will assume the inductance changes to L': \(f_0 = \dfrac{1}{2 \pi \sqrt{L' (2C)}}\)
02

Equate the two resonance frequency formulas

Since the resonance frequency remains unchanged, we can equate the two formulas: \(\dfrac{1}{2 \pi \sqrt{L C}} = \dfrac{1}{2 \pi \sqrt{L' (2C)}}\)
03

Solve for L'

Now we need to solve for L' in terms of L. First, we can cancel out constants on both sides of the equation: \(\sqrt{L C} = \sqrt{L' (2C)}\) Now square both sides to get rid of the square roots: \(L C = L' (2C)\) Since C ≠ 0, we can divide both sides by C: \(L = 2 L'\) Finally, we solve for L': \(L' = \dfrac{L}{2}\) So the inductance should change from L to L/2 for the resonance frequency to remain unchanged. The correct answer is (c) L/2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A transformer of efficiency \(90 \%\) draws an input power of \(4 \mathrm{~kW}\). An electrical appliance connected across the secondary draws a current of $6 \mathrm{~A}$. The impedance of device is ....... (a) \(60 \Omega\) (b) \(50 \Omega\) (c) \(80 \Omega\) (d) \(100 \Omega\)

The power factor of an ac circuit having resistance \((\mathrm{R})\) and inductance (L) connected in series and an angular velocity w is, (a) \((\mathrm{R} / \mathrm{cL})\) (b) $\left[\mathrm{R} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}^{(1 / 2)}\right]$ (c) \((\omega L / R)\) (d) $\left[\mathrm{R} /\left\\{\mathrm{R}^{2}-\omega^{2} \mathrm{~L}^{2}\right\\}^{(1 / 2)}\right]$

A wire of length \(2 \mathrm{~m}\) is moving at a speed \(2 \mathrm{~ms}^{-1}\) keep its length perpendicular to uniform magnetic field of \(0.5 \mathrm{~T}\). The resistance of circuit joined with this wire is \(6 \Omega\). The rate at which work is being done to keep the wire moving at constant speed is ........... (a) \((1 / 3) \infty\) (b) \((2 / 3) \omega\) (c) \((1 / 6) \propto\) (d) \(2 \mathrm{c}\)

Two coils of self inductances \(2 \mathrm{mH} \& 8 \mathrm{mH}\) are placed so close together that the effective flux in one coil is completely half with the other. The mutual inductance between these coils is...... (a) \(4 \mathrm{mH}\) (b) \(6 \mathrm{mH}\) (c) \(2 \mathrm{mH}\) (d) \(16 \mathrm{mH}\)

A resistor \(30 \Omega\), inductor of reactance \(10 \Omega\) and the capacitor of reactance \(10 \Omega\) are connected in series to an ac voltage source \(\mathrm{e}=300 \sqrt{2} \sin (\omega \mathrm{t})\) The current in the circuit is (a) \(10 \sqrt{2 \mathrm{~A}}\) (b) \(10 \mathrm{~A}\) (c) \(30 \sqrt{11 \mathrm{~A}}\) (d) \((30 / \sqrt{11}) \mathrm{A}\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free