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 $RC=\frac{2}{\omega }$
$=\frac{2}{500}s$
$=0.4\times 10^{-2}s=4$ millisecond
$\therefore$ The time constant of $RC$ circuit, $\tau =RC=4ms$