Question

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

b. What is V_R at this frequency ?

Short answer

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

(b) The expression for voltage across the resistor is $V_R=\frac{\sqrt{3}}{2}{\epsilon }_0$

Step-by-step solution

Part (a) Step 1 : Given information 

We are given that for an RC circuit, the voltage across a capacitor $V_C=\frac{1}{2}{\epsilon }_0$and

we have to find out the angular frequency$\omega$.

Part (a) Step 2 : Simplification 

As the expression for impedance of the circuit is,

$R=\frac{\sqrt{3}}{2}Z$

On rearranging we get,

$Z=\frac{\sqrt{3}}{2}R$

Now, the expression for angular frequency is,

$Z^2=R^2+\frac{1}{{\left(\omega C\right)}^2}$

Put the value of Z,

${\left(\frac{2R}{\sqrt{3}}\right)}^2=R^2+\frac{1}{{\omega }^2{C}^2}$

Simplify it,$\frac{1}{{\omega }^2{C}^2}=\left(\frac{4}{3}-1\right)R^2$

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

On solving we get,

$\omega =\frac{\sqrt{3}}{RC}$

Part (b) Step 1: Given Information

We are given that the voltage across a capacitor$V_c=\frac{1}{2}{\epsilon }_0$and

we have to find out the $V_R$i.e., potential across resistor.

Part (b) Step 2: Simplification

As we know,

The expression for voltage across resistor is,

${\epsilon }_0^2=V_R^2+V_C^2$

Put the value of$V_C$

${\epsilon }_0^2=V_R^2+{\left(\frac{{\epsilon }_0}{2}\right)}^2$

On simplifying,$V_R^2=\frac{3}{4}{\epsilon }_0^2$

We get$V_R=\frac{\sqrt{3}}{2}{\epsilon }_0$