Question 283: AC fundamentals

Question

An alternating e.m.f. having voltage $\mathrm{V}=\mathrm{V}_0 \sin \omega t$ is applied to a series L-C-R circuit. Given : $\left|X_L-X_C\right|=R$. The r.m.s. value of potential difference across capacitor will be

Accepted Answer

We are given an LCR
series circuit
with applied alternating emf:
$ V = V_0 \sin \omega t. $
Condition:
$ |X_L - X_C| = R $
where $X_L = \omega L$ and $X_C = 	frac{1}{\omega C}$.
We want the
r.m.s. value of potential difference across capacitor.
Step 1: Circuit impedance
In series circuit:
$ Z = \sqrt{R^2 + (X_L - X_C)^2}. $
Given condition: $|X_L - X_C| = R$.
So:
$ Z = \sqrt{R^2 + R^2} = \sqrt{2}\,R. $
Step 2: Current in circuit
Peak current:
$ I_0 = \frac{V_0}{Z} = \frac{V_0}{\sqrt{2}\,R}. $
r.m.s. current:
$ I_{	ext{rms}} = \frac{I_0}{\sqrt{2}} = \frac{V_0}{2R}. $
Step 3: Potential difference across capacitor
Voltage across capacitor (rms) is:
$ V_{C,	ext{rms}} = I_{	ext{rms}} \cdot X_C. $
So:
$ V_{C,	ext{rms}} = \frac{V_0}{2R} \cdot X_C. $
Now $X_C = \frac{1}{\omega C}$.
Thus:
$ V_{C,	ext{rms}} = \frac{V_0}{2R} \cdot \frac{1}{\omega C} = \frac{V_0}{2R\omega C}. $
Ã¢ÂÂ Final Answer:
The correct option is:
$ \boxed{	ext{Option C: } 	frac{V_0}{2R\omega C}} $

Answer

$\quad \mathrm{V}_0 \mathrm{R} \omega \mathrm{C}$

Answer

$\frac{\mathrm{V}_0}{\mathrm{R} \omega \mathrm{C}}$

Answer

$\frac{\mathrm{V}_0}{2 \mathrm{R} \omega \mathrm{C}}$

Answer

$\frac{\mathrm{V}_0}{\sqrt{2} \mathrm{R} \omega \mathrm{C}}$

AC fundamentals

An alternating e.m.f. having voltage \mathrm{V}=\mathrm{V}_0 \sin \omega t is applied to a series L-C-R circuit. Given : \left|X_L-X_C\right|=R. The r.m.s. value of potential difference across capacitor will be

Choose the correct answer: