At the start, the capacitor was uncharged. When switch is closed and is kept open, the time constant of this circuit is . Which of the following is correct? [2006]

Question

In the given circuit, the capacitor (C) may be charged through resistance R by a battery V by closing switch $S_{1}$ . Also, when $S_{1}$ is opened and $S_{2}$ is closed, the capacitor is connected in series with inductor (L).

Q. At the start, the capacitor was uncharged. When switch $S_{1}$ is closed and $S_{2}$ is kept open, the time constant of this circuit is $τ$ . Which of the following is correct? [2006]

Smart Achievers · Accepted Answer

(2)

When charging is complete, the potential difference between the capacitor plates will be V, and the charge stored in this case will be maximum.

$∴$ $S_{2}$

When $t = 2 τ,$

$τ$

$= C V (1 - e^{- 2})$

Solution

1 After time interval $τ$ , charge on the capacitor is $\frac{C V}{2}$

2 After time interval $2 τ$ , charge on the capacitor is $C V (1 - e^{- 2})$

3 The work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.

4 After time interval $2 τ$ , charge on the capacitor is $C V (1 - e^{- 1})$

Ans.
(2)

When charging is complete, the potential difference between the capacitor plates will be V, and the charge stored in this case will be maximum.

$∴$ $Q_{0} = C V$

When $t = 2 τ,$

$Q = C V [1 - e^{- \frac{2 τ}{τ}}]$

$= C V (1 - e^{- 2})$

Want Us to Email you About Special Offers & Updates?

Solution

1 After time interval τ, charge on the capacitor is CV2

2 After time interval 2τ, charge on the capacitor is CV(1-e-2)