
If the inductance and capacitance are both doubled in an L-C-R circuit, the resonant frequency of the circuit will:
A)Decrease to one-half of the original value
B) Decrease to one-fourth of the original value
C) Increase to twice the original value
D) Increase to thrice the original value
Answer
232.8k+ views
Hint: The output of the circuit is maximum at one particular frequency, that frequency is known as the resonant frequency of that electrical circuit. At resonant frequency, the capacitive reactance and inductive reactance are equal.
Formula used:
The resonant frequency of an L-C-R circuit is given by $\omega = \dfrac{1}{{\sqrt {LC} }}$, where L= inductance of the circuit, and C=capacitance of the circuit
Complete step by step solution:
First, let’s assume initial resonant frequency, inductance, and capacitance as ${\omega _o},{L_o},{C_o}$respectively.
Now after the operation resonant frequency, inductance, and capacitance changes to${\omega _{req}},{L_{req}},{C_{req}}$.
We know that
${L_{req}} = 2{L_o}$and${C_{req}} = 2{C_o}$
We also know that
${\omega _{req}} = \dfrac{1}{{\sqrt {{L_{req}}{C_{req}}} }}$
So, now by substituting these values in the formula of the required
resonant frequency(${\omega _{req}}$). We will get
${\omega _{req}} = \dfrac{1}{{\sqrt {\left( {2{L_o}} \right)\left( {2{C_o}} \right)} }}$
$ \Rightarrow {\omega _{req}} = \dfrac{1}{{2\sqrt {{L_o}{C_o}} }}$
$ \Rightarrow {\omega _{req}} = \dfrac{1}{2}{\omega _o}$ as ${\omega _o} = \dfrac{1}{{\sqrt {{L_o}{C_o}} }}$.
Hence we can say that resonant frequency is reduced to half of the initial value.
Additional information: In a parallel L-C-R circuit, the potential difference across the capacitor and the inductor are the same in magnitude but it is opposite in direction at the resonant frequency. As no power is consumed by the conductor and the capacitor and potential difference across the resistance is maximum, it results in the maximum power generated by the circuit.
Hence, the correct option is (A), Decrease to one-half of the original value.
Note:
1. The Resonant frequency of the circuit is independent of the resistance of the electrical circuit.
2. At the resonant frequency amplitude of the current is maximum in the electrical circuit.
3. Resonant frequency of the electrical circuit is inversely proportional to the square root of inductance and capacitance.
Formula used:
The resonant frequency of an L-C-R circuit is given by $\omega = \dfrac{1}{{\sqrt {LC} }}$, where L= inductance of the circuit, and C=capacitance of the circuit
Complete step by step solution:
First, let’s assume initial resonant frequency, inductance, and capacitance as ${\omega _o},{L_o},{C_o}$respectively.
Now after the operation resonant frequency, inductance, and capacitance changes to${\omega _{req}},{L_{req}},{C_{req}}$.
We know that
${L_{req}} = 2{L_o}$and${C_{req}} = 2{C_o}$
We also know that
${\omega _{req}} = \dfrac{1}{{\sqrt {{L_{req}}{C_{req}}} }}$
So, now by substituting these values in the formula of the required
resonant frequency(${\omega _{req}}$). We will get
${\omega _{req}} = \dfrac{1}{{\sqrt {\left( {2{L_o}} \right)\left( {2{C_o}} \right)} }}$
$ \Rightarrow {\omega _{req}} = \dfrac{1}{{2\sqrt {{L_o}{C_o}} }}$
$ \Rightarrow {\omega _{req}} = \dfrac{1}{2}{\omega _o}$ as ${\omega _o} = \dfrac{1}{{\sqrt {{L_o}{C_o}} }}$.
Hence we can say that resonant frequency is reduced to half of the initial value.
Additional information: In a parallel L-C-R circuit, the potential difference across the capacitor and the inductor are the same in magnitude but it is opposite in direction at the resonant frequency. As no power is consumed by the conductor and the capacitor and potential difference across the resistance is maximum, it results in the maximum power generated by the circuit.
Hence, the correct option is (A), Decrease to one-half of the original value.
Note:
1. The Resonant frequency of the circuit is independent of the resistance of the electrical circuit.
2. At the resonant frequency amplitude of the current is maximum in the electrical circuit.
3. Resonant frequency of the electrical circuit is inversely proportional to the square root of inductance and capacitance.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

