Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.96 cm^2, 2.02 cm^2, 2.90 cm^2, and 5.06 cm^2. A potential difference of 116 V is applied across the combination. Determine the voltage across the 2.02 cm^2 wire.

To determine the voltage across the 2.02 cm^2 wire, we can use the concept of resistance and Ohm's Law.

First, we need to calculate the resistance of each wire. The resistance (R) of a wire is given by the formula:

R = ρ * (L/A)

where ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.

Since all four copper wires are of the same length, the length (L) cancels out when calculating the ratio of their resistances. Therefore, we only need to calculate the ratio of their cross-sectional areas.

Let's label the wires as:

Wire 1: A1 = 0.96 cm^2
Wire 2 (2.02 cm^2): A2 = 2.02 cm^2
Wire 3: A3 = 2.90 cm^2
Wire 4: A4 = 5.06 cm^2

The ratio of the cross-sectional areas can be calculated as:

A2/A1 = 2.02 cm^2 / 0.96 cm^2 = 2.10

Now, we can calculate the voltage across the 2.02 cm^2 wire using the concept of potential division. According to potential division, the voltage across a specific resistor in a series circuit is proportional to the ratio of its resistance to the total resistance in the circuit.

Let V be the voltage across the 2.02 cm^2 wire. The ratio of the resistance of the 2.02 cm^2 wire to the total resistance is equal to the ratio of the voltage across the 2.02 cm^2 wire to the total voltage.

Therefore, we have:

R2 / (R1 + R2 + R3 + R4) = V / (V + 116 V)

Since the lengths of all wires are equal and the resistivity of copper is the same for all wires, the resistivity (ρ) cancels out. We are left with:

A2 / (A1 + A2 + A3 + A4) = V / (V + 116 V)

Substituting the values, we get:

2.02 cm^2 / (0.96 cm^2 + 2.02 cm^2 + 2.90 cm^2 + 5.06 cm^2) = V / (V + 116 V)

Simplifying the equation:

2.02 cm^2 / 10.94 cm^2 = V / (V + 116 V)

Cross-multiplying:

2.02 cm^2 * (V + 116 V) = 10.94 cm^2 * V

2.02V + 233.52V = 10.94V

235.54V = 10.94V

Now, we can solve for V:

235.54V - 10.94V = 0

224.60V = 0

V = 0V

Therefore, the voltage across the 2.02 cm^2 wire is 0 volts.