Question

In the circuit shown, when the switch is closed, the capacitor
charges with a time constant
(a) RC
(b) 2RC
(c) 1/2 RC
(d) RC ln 2

Answer 1

Given that,

Capacitor = C

Voltage = V

Charge = q

According to figure,

The current in second loop is

$i_{2}=\dfrac{V}{R}$

Where, V = voltage

R = resistance

We know that,

The current is

$i=\dfrac{dq}{dt}$.....(I)

We need to calculate the charge

Using Kirchhoff's law in second loop

$V-\dfrac{q}{C}-iR=0$

$\dfrac{CV-q}{C}=iR$

Pit the value of i from equation (I)

$\dfrac{CV-q}{C}=\dfrac{dq}{dt}R$

$\dfrac{dq}{CV-q}=\dfrac{dt}{RC}$

On integration

$\int_{0}^{q}{\dfrac{dq}{CV-q}}=\int_{0}^{t}{\dfrac{dt}{RC}}$

$-ln(\dfrac{CV-q}{CV})=\dfrac{t}{RC}$

$q=CV(1-e^{-\frac{t}{RC}})$

$q=CV(1-e^{-\frac{t}{\tau}})$

Here, $\tau = RC$

Hence, The capacitor  charges with a time constant is RC.

(a) is correct option.