In a circuit shown in the figure, the capacitor C is initially...

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

Now for the first branch, we can write

$15 - IR = 6 . . . \left(1\right)$ and from third branch, we can write

${I}_{1} \times 3 = 6$

$⇒ {I}_{1} = 2A$

Now from Kirchoff's junction rule,

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

Now, from equation(1), we can write

$15 - 3R = 6$

$⇒ R = 3Ω$

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

$\dfrac{1}{{R}_{eq}} = \dfrac{1}{3} + \dfrac{1}{3} + 1$

$⇒ {R}_{eq} = \dfrac{3}{5}Ω = 0 . 6Ω$

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

${E}_{eq} = 6V$.

Now, current variation in the circuit due to charging will be $\dfrac{6}{\left(\dfrac{3}{5} + 3\right)}e\dfrac{- \left(t - {t}_{0}\right)}{CR}$

At, $t = {t}_{0} + 7 . 2\mu s$, we get

$i = \dfrac{6 \times 5}{18}{e}^{- \dfrac{\left(7 . 2 \times {10}^{- 6}\right)}{\left(2 \times {10}^{- 6}\right) \times \left(3 . 6\right)}}$

$= \dfrac{30}{18} \times {e}^{- 1} = \dfrac{30}{18} \times 0 . 36$

$= 0 . 6A$

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

$Q = 6 \times 2 = 12\muC$.