
In an L-C-R circuit the value of ${X_L}$ , ${X_C}$ and $R$ are $300\Omega $ , $200\Omega $ and $100\Omega $ respectively. The total impedance of the circuit will be
(A) $600\Omega $
(B) $200\Omega $
(C) $141\Omega $
(D) $310\Omega $
Answer
141k+ views
Hint: - At resonance the capacitive reactance and inductive reactance are equal. At above the resonant frequency the inductive reactance will be greater than the capacitive reactance. So the RLC circuit operating above resonant frequency behaves as a purely inductive circuit. The phase difference between the current and voltage will be equal in this circuit as the inductive circuit.
Formula used:
The formula for the impedance in the LCR circuit,
$Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $
where $Z$ = impedance
$R$ = resistance
${X_L}$ = Inductive reactance
${X_C}$ = Capacitive Reactance
Complete step-by-step solution:
Given,
The value of the resistor in the RLC circuit is $100\Omega $ ,
The value of Inductive reactance in the RLC circuit is $300\Omega $ ,
The value of Capacitive reactance in the RLC circuit is $200\Omega $ .
Since, For a series LCR circuit, the impedance is given by the formula,
$Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $
Substitute the value of $R$ , ${X_L}$ and ${X_C}$ in the above equation we get,
$Z = \sqrt {{{\left( {100} \right)}^2} + {{\left( {300 - 200} \right)}^2}} $
$ \Rightarrow Z = \sqrt {{{\left( {100} \right)}^2} + {{\left( {100} \right)}^2}} $
On further solving the equation we get,
$Z = \sqrt {2{{\left( {100} \right)}^2}} = 100\sqrt 2 $
$ \Rightarrow Z = 141.42 \simeq 141\Omega $
And so the total impedance of the circuit is $141\Omega $ .
Hence, the correct answer is option is (C).
Additional information: In the LCR circuits when the ${X_L} \succ {X_C}$ , then the circuit is termed as Inductive circuit and similarly when the ${X_C} \succ {X_L}$ then the corresponding circuits are termed as Capacitive circuit. Always remember that the voltages in a series RLC circuit are actually phasors which are treated as vectors, so the net emf is obtained as a vector addition of three voltages given.
Note: The series resonance or the series LCR circuits are one of the most significant circuits. They have a vast number of practical uses starting from AC mains filters, radios, and also in television circuits. We have to remember all the formulas used and understand the meaning of the terms such as impedance and reactance, then this type of question will be solved easily.
Formula used:
The formula for the impedance in the LCR circuit,
$Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $
where $Z$ = impedance
$R$ = resistance
${X_L}$ = Inductive reactance
${X_C}$ = Capacitive Reactance
Complete step-by-step solution:
Given,
The value of the resistor in the RLC circuit is $100\Omega $ ,
The value of Inductive reactance in the RLC circuit is $300\Omega $ ,
The value of Capacitive reactance in the RLC circuit is $200\Omega $ .
Since, For a series LCR circuit, the impedance is given by the formula,
$Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $
Substitute the value of $R$ , ${X_L}$ and ${X_C}$ in the above equation we get,
$Z = \sqrt {{{\left( {100} \right)}^2} + {{\left( {300 - 200} \right)}^2}} $
$ \Rightarrow Z = \sqrt {{{\left( {100} \right)}^2} + {{\left( {100} \right)}^2}} $
On further solving the equation we get,
$Z = \sqrt {2{{\left( {100} \right)}^2}} = 100\sqrt 2 $
$ \Rightarrow Z = 141.42 \simeq 141\Omega $
And so the total impedance of the circuit is $141\Omega $ .
Hence, the correct answer is option is (C).
Additional information: In the LCR circuits when the ${X_L} \succ {X_C}$ , then the circuit is termed as Inductive circuit and similarly when the ${X_C} \succ {X_L}$ then the corresponding circuits are termed as Capacitive circuit. Always remember that the voltages in a series RLC circuit are actually phasors which are treated as vectors, so the net emf is obtained as a vector addition of three voltages given.
Note: The series resonance or the series LCR circuits are one of the most significant circuits. They have a vast number of practical uses starting from AC mains filters, radios, and also in television circuits. We have to remember all the formulas used and understand the meaning of the terms such as impedance and reactance, then this type of question will be solved easily.
Recently Updated Pages
Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
