Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Given : $\omega =500$ $radian/s$

Let the capacitance of the capacitor be $C$.

Thus resistance of capacitor $X_C=\frac{1}{\omega C}=\frac{1}{500C}$

Impedance of the circuit $Z=R\sqrt{1.25}$

Using $Z^2=R^2+X_C^2$

$\therefore$ $1.25R^2=R^2+\frac{1}{(500{)}^2{C}^2}$

Or $0.25R^2=\frac{1}{.25\times {10}^6{C}^2}$

$⟹$ $R^2{C}^2=\frac{{10}^{-6}}{(0.25{)}^2}$

We get time constant of the circuit $RC=\frac{{10}^{-3}}{0.25}=0.004$ s $=4$ ms