(4)

[IMAGE 192]---------------------------------------

Let potential at junction $C$ be $x$

        $\frac{x-50}{1}+\frac{x-50}{3}+\frac{x-0}{2}+\frac{x-0}{4}=0$

        $12x-600+4x-200+6x+3x=0$

         $25x-800=0$

or      $x=\frac{800}{25}=32\text{V}$

Current through $1\Omega$ resistance

         $I_1=\frac{50-32}{1}=18$$A$

Current through $2\Omega$ resistance

        $I_2=\frac{32-0}{2}=16\text{A}$

$\therefore$   Current through branch CD is

        $I_{CD}=I_1-I_2=18-16=2\text{A}$

(3)

In steady state capacitor acts as an open circuit.

[IMAGE 194]---------------

So, the circuit will become

$R_{\text{eq}}=2+3=5\Omega$

$V=10\text{V}$

$I=\frac{V}{R_{\text{eq}}}=\frac{10}{5}=2$$\text{A}$

(1)

Since galvanometer shows no deflection, thus voltage across $R$ is $2V$.

Now, voltage across $R$ is, $V=\frac{10R}{400+R}$

or    $2=\frac{10R}{400+R} \Rightarrow R=100\Omega$

(4)

[IMAGE 197]

$V_{AB}=V_A-V_B=2\times 2+3+1\times 2=9$ $\text{V}$