Question

In the series $LCR$ circuit at resonance, the applied ac voltage is $220V$. The potential drop across the inductance is $110V$. What is the potential drop across the resistance?

Answer 1

Step 1: Given Data

The applied ac voltage $=220V$

The potential drop across the inductance $=110V$

Step 2: Series $LCR$ Circuit at Resonance

We know that the current in a series $LCR$ circuit impressed by an ac emf $v=V_0\cos \omega t$ is given by,

$I=I_0\cos \left(\omega t-\varphi \right)$,

Where, $I_0=\frac{V_0}{\sqrt{R^2+{\left(\omega L-{\displaystyle \frac{1}{\omega C}}\right)}^2}}$ and $\varphi ={\tan }^{-1}\frac{\omega L-{\displaystyle \frac{1}{\omega C}}}{R}$

The RMS current is $I_r=\frac{V_r}{\sqrt{R^2+{\left(\omega L-{\displaystyle \frac{1}{\omega C}}\right)}^2}}$, $\left(1\right)$

Where $V_r$ is the RMS voltage. Thus, $I_r$ depends on the frequency $\omega$ of the impressed ac emf. For a certain value of $\omega$, $I_r$ becomes maximum and we say there is resonance.

In this case, $\omega L=\frac{1}{\omega C}$.

Hence, equation $\left(1\right)$ becomes,

$I_r=\frac{V_r}{\sqrt{R^2+{\left(0\right)}^2}}$

$\Rightarrow I_r=\frac{V_r}{R}$ $\left(2\right)$

Step 3: Potential drop calculation

From the question, it is given that the circuit is at resonance.

When the circuit is at resonance, $\omega L-\frac{1}{\omega C}=0$, which is why the voltage drop is only utilized by the resistance.

Hence, here we will use equation $\left(2\right)$.

The equation $\left(2\right)$ is nothing but the form of Ohm's law.

Therefore, $V_r=V=220V$

Hence, the potential drop across the resistance is $220V$,