Question

a. For an RC circuit, find an expression for the angular frequency at which $V_R=\frac{1}{2}{\epsilon }_0$

b. What is V_C at this frequency?

Short answer

(a) The final expression for angular frequency is $\omega =\frac{1}{\sqrt{3}RC}$

(b) The V_C at the given frequency is$V_C=\frac{\sqrt{3}{\epsilon }_0}{2}$

Step-by-step solution

Part (a) Step 1: Given Information

We are given that for an RC circuit, the voltage across resistor is $V_R=\frac{1}{2}{\epsilon }_0$.

We have to find out the expression for angular frequency.

Part (a) Step 2: Simplification

We know,

$V_R=\frac{{\epsilon }_{0}R}{\sqrt{R^2+{\displaystyle \frac{1}{{\omega }^2{c}^2}}}}$ (where $V_R=\frac{1}{2}{\epsilon }_0$)

Here $V_R$is voltage across resistor, R is the resistor, $\omega$is the angular frequency and $c$is the capacitor.

On simplifying we get,$\sqrt{1+\frac{1}{{\omega }^2{C}^2{R}^2}}=2$

$\frac{1}{{\omega }^2{C}^2{R}^2}=3$

which gives the result ,$\omega =\frac{1}{\sqrt{3}RC}$

Part (b) Step 1: Given Information

We are given that for an RC circuit, the voltage across the resistor is $V_R=\frac{1}{2}{\epsilon }_0$and we have already find out the angular frequency $\omega =\frac{1}{\sqrt{3}RC}$.

We have to find out the$V_C$.

Part (b) Step 2: Simplification

We know, the formula for voltage across the capacitor is

$V_C=\frac{{\displaystyle \frac{{\epsilon }_0}{{\omega }_C}}}{\sqrt{R^2+{\displaystyle \frac{1}{{\omega }^2{C}^2}}}}$

Substituting all the values in it, we get

$V_C=\frac{\sqrt{3}{\epsilon }_0}{2}$.

The voltage across capacitor$V_C=\frac{\sqrt{3}{\epsilon }_o}{2}$