Question 341: AC fundamentals

Question

A series LCR circuit containing a resistance ' R ' has angular frequency ' $\omega$ '. At resonance the voltage across resistance and inductor are ' $\mathrm{V}_{\mathrm{R}}$ ' and ' $V_L$ ' respectively, then value of inductance ' $L$ ' will be

Accepted Answer

In a series LCR circuit at resonance, the impedance is purely resistive, and the resonance condition is achieved when the reactive components of the impedance, i.e., the inductive reactance ($X_L = \omega L$) and capacitive reactance ($X_C = \frac{1}{\omega C}$), are equal.
At resonance, the voltage across the inductor can be related to the voltage across the resistor using their reactance and resistance. The voltage across the inductor $V_L$ can be expressed as:
$$ V_L = I \cdot \omega L $$
And the voltage across the resistor $V_R$ is:
$$ V_R = I \cdot R $$
where $I$ is the current through the circuit.
Since at resonance the same current $I$ flows through both the resistor and the inductor:
$$ I = \frac{V_R}{R} $$
Substitute $I$ into the expression for $V_L$:
$$ V_L = \frac{V_R}{R} \cdot \omega L $$
From this, rearrange to find the inductance $L$:
$$ L = \frac{V_L R}{V_R \omega} $$
Thus, the correct option is
Option D
:
$$ \frac{\mathrm{V}_{\mathrm{L}} \mathrm{R}}{\mathrm{V}_{\mathrm{R}} \omega} $$

Answer

$\frac{V_R R}{V_L \omega}$

Answer

$\frac{V_L}{V_R R \omega}$

Answer

$\frac{V_R \omega}{V_L R}$

Answer

$\frac{\mathrm{V}_{\mathrm{L}} \mathrm{R}}{\mathrm{V}_{\mathrm{R}} \omega}$

AC fundamentals

A series LCR circuit containing a resistance ' R ' has angular frequency ' \omega '. At resonance the voltage across resistance and inductor are ' \mathrm{V}_{\mathrm{R}} ' and ' V_L ' respectively, then value of inductance ' L ' will be

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