Question

Time constant of series $RC$ circuit is$\tau $. If the frequency of applied a.c. is $\omega = \dfrac{1}{\tau }$, find the impedance of the circuit.
a. $\sqrt 2 R$
b. $2R$
c. $\dfrac{R}{{\sqrt 2 }}$
d. $\dfrac{R}{2}$

Answer 1

Hint: The time constant of an $RC$ circuit is equal to the product of capacitance and resistance. The impedance of a circuit is given as the square root of the sum of its resistance and capacitive reactance squared. The given value of frequency can be used to calculate the capacitive reactance of the circuit, and the impedance can be calculated.

Formula used:
$\tau = RC$
$Z = \sqrt {{R^2} + {X_C}^2} $
where $\tau $ is the time constant,
$R$ is the resistance in the circuit,
$C$ is the capacitance of the circuit,
$Z$ is the impedance of the circuit,
${X_C}$ is the capacitive reactance of the circuit.

Complete step by step answer:
The time constant of an $RC$ circuit is the time taken to charge the capacitor to $63.2\% $ of the total voltage applied across it. It is given by the formula-
$\tau = RC$

The capacitive reactance of a circuit is used to donate the impedance or the resistance to current flow by the capacitive component of that circuit. It is given by,
${X_C} = \dfrac{1}{{\omega C}}$
Here $\omega $ is the frequency of a.c. current and $C$ is the capacitance of the capacitor.
In the question, it is given that-
$\omega = \dfrac{1}{\tau }$
By substituting this value of $\omega $, we get-
${X_C} = \dfrac{\tau }{C}$

Substituting the value of $\tau $, we get-
${X_C} = \dfrac{{RC}}{C}$
$ \Rightarrow {X_C} = R$
We know that the impedance of an $RC$ circuit is given by-
$Z = \sqrt {{R^2} + {X_C}^2} $
Substituting the value of ${X_C}$ in this formula, we get-
$Z = \sqrt {{R^2} + {R^2}} $
$ \Rightarrow Z = \sqrt 2 R$
The impedance of the circuit is $\sqrt 2 R$.

Hence, the correct answer is option (A).

Note: The time constant is an exponential term, if a capacitor takes $\tau $ seconds to charge to $63.2\% $, it will take another $\tau $ seconds to reach top $86.5\% $ of the voltage applied at it. It signifies that the rate of charging of a capacitor slows down exponentially as it gets charged. Thus for a capacitor to get charged to $100\% $of the applied voltage, the time required would be infinite, this is a state known as the steady-state.