A constant voltage of 50 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 find the current through the branch CD of the circuit, we can use the nodal analysis method.

Let us analyze the circuit shown in the diagram:
- A constant voltage of 50 V is maintained between terminals A and B. Let the electric potential at terminal B be $V_B=0\text{V}$, which means the electric potential at terminal A is $V_A=50\text{V}$.
- The branch CD is a connecting wire with zero resistance. This means the nodes C and D are at the same electric potential. Let this common potential be $V_C=V_D=V$.

Now, let's write the expression for the currents entering and leaving the combined junction (supernode) C-D:

1. Current flowing from A to C through the 1 Ω resistor:
$I_{\mathrm{AC}}=\frac{V_A-V_C}{1}=\frac{50-V}{1}$

2. Current flowing from A to D through the 3 Ω resistor:
$I_{\mathrm{AD}}=\frac{V_A-V_D}{3}=\frac{50-V}{3}$

3. Current flowing from C to B through the 2 Ω resistor:
$I_{\mathrm{CB}}=\frac{V_C-V_B}{2}=\frac{V}{2}$

4. Current flowing from D to B through the 4 Ω resistor:
$I_{\mathrm{DB}}=\frac{V_D-V_B}{4}=\frac{V}{4}$

Applying Kirchhoff's Current Law (KCL) to the combined C-D supernode (Total current entering = Total current leaving):

$I_{\mathrm{AC}}+I_{\mathrm{AD}}=I_{\mathrm{CB}}+I_{\mathrm{DB}}$

Substituting the expressions into the KCL equation:

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

Multiply the entire equation by 12 (the least common multiple of 2, 3, and 4) to eliminate the denominators:

$12(50-V)+4(50-V)=6V+3V$

Combine like terms:

$16(50-V)=9V$

$800-16V=9V$

$25V=800$

$V=\frac{800}{25}=32\text{V}$

Thus, the potential at node C and node D is $V=32\text{V}$.

Next, we calculate the individual currents at node C:
- Current entering node C from terminal A:
$I_{\mathrm{AC}}=50-32=18\text{A}$
- Current leaving node C to terminal B:
$I_{\mathrm{CB}}=\frac{32}{2}=16\text{A}$

By applying Kirchhoff's Current Law at node C, the net current must balance. Since 18 A enters from A and only 16 A leaves towards B, the remaining current must flow downwards from node C to node D through the branch CD:

$I_{\mathrm{CD}}=I_{\mathrm{AC}}-I_{\mathrm{CB}}=18\text{A}-16\text{A}=2.0\text{A}$

Alternatively, verifying at node D:
- Current entering node D from terminal A:
$I_{\mathrm{AD}}=\frac{50-32\}}{3}=6\text{A}$
- Current leaving node D to terminal B:
$I_{\mathrm{DB}}=\frac{32}{4}=8\text{A}$
Since 8 A leaves towards B but only 6 A enters from A, the additional 2.0 A must enter node D from branch CD, confirming a downward current of 2.0 A.

Therefore, the current through the branch CD is 2.0 A.