Figure 9 shows some of the apparatus used in a demonstration of electrical power transmission using a dc power supply - AQA - A-Level Physics - Question 5 - 2020 - Paper 1

The total current from the power supply can be calculated using the total resistance in the circuit:

Since the lamps are in parallel, the total resistance R_total is given by:

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

where each lamp has a resistance of about 100 Ω:

$\frac{1}{R_{total}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{100} = \frac{3}{100}$

Thus,

$R_{total} = \frac{100}{3} \approx 33.33\ Ω$

Applying Ohm's Law:

$I = \frac{V}{R_{total}} = \frac{12\ V}{33.33\ Ω} \approx 0.36\ A$

The resistance of a wire can be calculated using the formula:

$R = \rho \frac{L}{A}$

Where:

• $R$ is the resistance
• $\rho$ is the resistivity
• $L$ is the length of the wire
• $A$ is the cross-sectional area.

The area A of a cylindrical wire is given by:

$A = \pi r^2$

First, calculate the radius from the diameter:

$d = 0.19\ mm = 0.19 \times 10^{-3}\ m \implies r = \frac{d}{2} = \frac{0.19\times 10^{-3}}{2}\ m \approx 0.095\times 10^{-3}\ m$

Then,

$A = \pi (0.095\times 10^{-3})^2 \approx 2.84\times 10^{-9}\ m^2$

Now substituting into the resistance formula with $\rho = 4.9 \times 10^{-7}\ Omega m$, and $L = 2.8\ m$:

$R = 4.9 \times 10^{-7} \frac{2.8}{2.84 \times 10^{-9}} \approx 50\ Ω$

In the demonstration, when using the long constantan wires, the resistance increases significantly due to their length compared to the short copper wires, which have negligible resistance.

Thus, the total resistance in the circuit with the constantan wires will lead to a decrease in current flowing through the lamps.

To show the effect quantitatively, the resistance of the constantan wires might introduce: $R_{total} = R_{lamps} + R_{constantan}$.

Where the increase in resistance results in using: $I = \frac{12\ V}{R_{total}}$.

This decrease in current will subsequently mean that the lamps will receive less power and appear dimmer when connected via the long constantan wires compared to the short copper wires, achieving the purpose of the demonstration.

One significant advantage of using superconductors is their zero resistivity, which allows for unlimited current flow without energy loss. This efficiency is crucial for long-distance electrical transmission, reducing energy costs and improving system performance.

However, a notable difficulty is that superconductors require extremely low temperatures to maintain their superconducting state, often involving complex and costly cooling technology. This temperature maintenance poses challenges for practical applications and can hinder widespread implementation in electrical transmission systems.