Question

A series $R$-$C$ combination is connected to an $AC$ voltage of angular frequency $\omega = 500$ radian/s. If the impedance of the $R$-$C$ circuit is $R\sqrt{1.25}$, the time constant (in millisecond) of the circuit is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (C)

Here, $\omega = 500$ radian/s

The capacitive reactance is
$X_{C}=\frac{1}{\omega C}$
The impedance of the circuit is
$Z=\sqrt{R^{2}+\left(X_{C}\right)^{2}}$
$=\sqrt{R^{2}+\left(\frac{1}{\omega C}\right)^{2}}$
$R\sqrt{\frac{5}{4}}=\sqrt{R^{2}+\left(\frac{1}{\omega C}\right)^{2}}$ or $\frac{5}{4}R^{2}=R^{2}+\left(\frac{1}{\omega C}\right)^{2}$
$\frac{1}{4}R^{2}=\left(\frac{1}{\omega C}\right)^{2}$ or $R^{2}C^{2}=4\left(\frac{1}{\omega}\right)^{2}$
or $RC=\frac{2}{\omega}$
$=\frac{2}{500}s$
$=0.4\times10^{-2}\,s=4$ millisecond
$\therefore $ The time constant of $RC$ circuit, $\tau = RC = 4 \,ms$