Question

A charged capacitor discharges through a resistance R with time constant $\tau$. The two are now placed in series across an AC source of angular frequency $\omega =\frac{1}{\tau }$ . The impedance of the circuit will be:

• A. $\frac{R}{\sqrt{2}}$
• B. R
• C. $\sqrt{2}$ R
• D. 2R

Answer 1

Answer: (C)

$\sqrt{2}$ R

In RC circuit, impedance $Z=\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}$ and time constant $\tau =CR$

Since $\omega =\frac{1}{\tau }$ so $Z=\sqrt{R^2+\frac{{\tau }^2}{C^2}}=\sqrt{R^2+\frac{C^2{R}^2}{C^2}}=\sqrt{2}R$