In a circuit shown in the figure, the capacitor $C$ is initially uncharged and the key $K$ is open. In this condition, a current of $1 A$ flows through the $1 Ω$ resistor. The key is closed at time $t = t_{0}$ . Which of the following statement(s) is(are) correct?
[Given : $e^{- 1} = 0.36$ ]

Question

In a circuit shown in the figure, the capacitor $C$ is initially uncharged and the key $K$ is open. In this condition, a current of $1 A$ flows through the $1 Ω$ resistor. The key is closed at time $t = t_{0}$ . Which of the following statement(s) is(are) correct?

[Given : $e^{- 1} = 0.36$ ]

Aicharya · Accepted Answer

From second branch, we can see the potential drop across the branch is $5 + (1 \times 1) = 6 V$ .

Now for the first branch, we can write

$15 - I R = 6 . . . (1)$ and from third branch, we can write

$I_{1} \times 3 = 6$

$\Rightarrow I_{1} = 2 A$

Now from Kirchoff's junction rule,

$I = I_{1} + 1 = 3$

Now, from equation(1), we can write

$15 - 3 R = 6$

$\Rightarrow R = 3 Ω$

All three branches are in parallel, therefore we can write equivalent resistance as:

$\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{3} + 1$

$\Rightarrow R_{eq} = \frac{3}{5} Ω = 0.6 Ω$

As all branches are in parallel with the capacitor branch, potential drop across all three branches will be the same. Therefore,

$E_{eq} = 6 V$ .

Now, current variation in the circuit due to charging will be $\frac{6}{(\frac{3}{5} + 3)} e^{\frac{- (t - t_{0})}{C R}}$

At, $t = t_{0} + 7.2 μs$ , we get

$i = \frac{6 \times 5}{18} e^{- \frac{(7.2 \times 10^{- 6})}{(2 \times 10^{- 6}) \times (3.6)}}$

$= \frac{30}{18} \times e^{- 1} = \frac{30}{18} \times 0.36$

$= 0.6 A$

At steady state, voltage across capacitor $= 6 V$ .

$Q = 6 \times 2 = 12 μ C$ .

In a circuit shown in the figure, the capacitor C is initially uncharged and the key K is open. In this…

Solution