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(C) Let the potential at the right junction be $V$ and the left junction be $0 \,V$.
For the branch with $1 \,\Omega$ resistor,the current is $1 \,A$

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(C) Let the potential at the right junction be $V$ and the left junction be $0 \,V$.
For the branch with $1 \,\Omega$ resistor,the current is $1 \,A$ flowing towards the left junction. Thus,$V - 5 = 1 	imes 1$,which gives $V = 6 \,V$.
For the branch with $R$,the current $I$ flows from left to right. Thus,$15 - I 	imes R = V = 6$,so $I 	imes R = 9$.
For the branch with $3 \,\Omega$ resistor,the current $I_1$ flows from right to left. Thus,$V - 3 	imes I_1 = 0$,which gives $6 - 3 	imes I_1 = 0$,so $I_1 = 2 \,A$.
At the right junction,by Kirchhoff's Current Law,$I = 1 + I_1 = 1 + 2 = 3 \,A$.
Since $I 	imes R = 9$,we have $3 	imes R = 9$,so $R = 3 \,\Omega$. Thus,$(A)$ and $(B)$ are correct.
When $K$ is closed,the circuit acts as a capacitor charging through an equivalent $EMF$ $\varepsilon_{eq}$ and equivalent resistance $r_{eq}$.
Using Millman's theorem,$\varepsilon_{eq} = \frac{\frac{15}{3} + \frac{5}{1} + \frac{0}{3}}{\frac{1}{3} + \frac{1}{1} + \frac{1}{3}} = \frac{5 + 5}{5/3} = 6 \,V$.
The equivalent resistance $r_{eq}$ is the parallel combination of $3 \,\Omega, 1 \,\Omega, 3 \,\Omega$,which is $\frac{1}{r_{eq}} = \frac{1}{3} + 1 + \frac{1}{3} = \frac{5}{3} \,\Omega^{-1}$,so $r_{eq} = 0.6 \,\Omega$.
The total resistance in the capacitor branch is $R_{total} = r_{eq} + 3 \,\Omega = 0.6 + 3 = 3.6 \,\Omega$.
The time constant $	au = R_{total} 	imes C = 3.6 \,\Omega 	imes 2 \,\mu F = 7.2 \,\mu s$.
The current in the capacitor branch is $i(t) = \frac{\varepsilon_{eq}}{R_{total}} e^{-t/	au} = \frac{6}{3.6} e^{-t/	au} = \frac{5}{3} e^{-t/	au}$.
At $t = t_0 + 7.2 \,\mu s$,$i = \frac{5}{3} e^{-1} = \frac{5}{3} 	imes 0.36 = 0.6 \,A$. Thus,$(C)$ is correct.
As $t < \infty$,the capacitor is fully charged,$q_{max} = C 	imes \varepsilon_{eq} = 2 \,\mu F 	imes 6 \,V = 12 \,\mu C$. Thus,$(D)$ is correct.

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