A constant voltage of $50\text{V}$ is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:

To solve for the current through branch CD, we need to find the current flowing from the top node to the bottom node across the central wire.1. Circuit AnalysisThe branch CD is a zero-resistance wire (a short circuit). This means points C and D are at the same electrical potential ($V_C=V_D$).Because they are at the same potential, we can treat the circuit as two parallel combinations connected in series:Left Section: The $1\Omega$ and $3\Omega$ resistors are in parallel.Right Section: The $2\Omega$ and $4\Omega$ resistors are in parallel.2. Calculate Resistance and Total CurrentFirst, find the equivalent resistance for each section:$R_{left}$: $\frac{1\times 3}{1+3}=\frac{3}{4}=0.75\Omega$$R_{right}$: $\frac{2\times 4}{2+4}=\frac{8}{6}=1.333\Omega$ Total Resistance ($R_{total}$):
$$R_{total}=0.75+1.333=2.083\Omega$$
Total Circuit Current ($I$):Using $V=50\text{ V}$:
$$I=\frac{50}{2.083}\approx 24\text{ A}$$
3. Calculate Branch CurrentsNow, find the current entering node C and node D from the left side using the current division rule:Current through $1\Omega$ ($I_1$): $I\times \frac{3}{1+3}=24\times 0.75=18\text{ A}$ Current through $3\Omega$ ($I_3$): $I\times \frac{1}{1+3}=24\times 0.25=6\text{ A}$ Next, find the current leaving node C and node D toward the right:Current through $2\Omega$ ($I_2$): $I\times \frac{4}{2+4}=24\times \frac{2}{3}=16\text{ A}$ Current through $4\Omega$ ($I_4$): $I\times \frac{2}{2+4}=24\times \frac{1}{3}=8\text{ A}$ 4. Apply Kirchhoff’s Current Law (KCL) at Node CAt node C, the current entering from the left must equal the current leaving to the right and through the wire CD.
$$I_1=I_2+I_{CD}$$
$$18\text{ A}=16\text{ A}+I_{CD}$$
$$I_{CD}=18-16=2.0\text{ A}$$
Correct Option: (2) 2.0 A