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:

The correct answer is 2.0 A.

From the circuit diagram, we can identify the following components and electrical nodes:
- A DC voltage source of 50 V maintains a constant potential difference between points A and B.
- A 1 Ω resistor is connected between A and C.
- A 3 Ω resistor is connected between A and D.
- A 2 Ω resistor is connected between C and B.
- A 4 Ω resistor is connected between D and B.
- Branch CD is a zero-resistance path connecting nodes C and D.

Since nodes C and D are directly connected by a conducting wire of zero resistance, they must be at the same electrical potential:
$V_C=V_D=V$

Let us define the reference potential at node B as $V_B=0\text{V}$. Consequently, the potential at node A is:
$V_A=50\text{V}$

Using Ohm's Law, we can express the current in each resistor in terms of the node potentials:
- Current through the 1 Ω resistor ($I_{AC}$):
$I_{AC}=\frac{50-V}{1}=50-V$
- Current through the 3 Ω resistor ($I_{AD}$):
$I_{AD}=\frac{50-V}{3}$
- Current through the 2 Ω resistor ($I_{CB}$):
$I_{CB}=\frac{V-0}{2}=\frac{V}{2}$
- Current through the 4 Ω resistor ($I_{DB}$):
$I_{DB}=\frac{V-0}{4}=\frac{V}{4}$

Applying Kirchhoff's Current Law (KCL) to the combined junction of C and D:
$I_{AC}+I_{AD}=I_{CB}+I_{DB}$

Substituting the current expressions:
$(50-V)+\frac{50-V}{3}=\frac{V}{2}+\frac{V}{4}$

Simplifying the equations:

$\frac{4}{3}(50-V)=\frac{3}{4}V$

Multiplying the entire equation by 12 to eliminate fractions:
$16(50-V)=9V$
$800-16V=9V$
$25V=800$
$V=32\text{V}$

Now, we calculate the individual currents at node C:
$I_{AC}=50-32=18\text{A}$
$I_{CB}=\frac{32}{2}=16\text{A}$

Applying Kirchhoff's Current Law specifically at node C:
$I_{AC}=I_{CB}+I_{CD}$
$18=16+I_{CD}$
$I_{CD}=2.0\text{A}$

Thus, the current through branch CD is 2.0 A.