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

Using Ohm's Law, the total current can be calculated from the equivalent resistance of the parallel circuit:

1. Calculate the equivalent resistance of three lamps in parallel:

R_{total} = rac{R}{3} = rac{100}{3} \approx 33.33 \, Ω

2. Calculate the total current using Ohm's law:

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

Thus, the current in the power supply is approximately 0.36 A.

To find the resistance of each length of constantan wire, we use the formula:

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

Where:

• $R$ is the resistance
• $\rho$ is the resistivity of constantan (4.9 × 10⁻⁷ Ω m)
• $L$ is the length (2.8 m)
• $A$ is the cross-sectional area.

Calculating the area for a wire of diameter 0.19 mm:

$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.19 \times 10^{-3}}{2}\right)^2 \approx 2.8 \times 10^{-8} \ m^2$

Now, substituting into the resistance formula:

$R = 4.9 \times 10^{-7} \cdot \frac{2.8}{2.8 \times 10^{-8}} \approx 50 \, Ω$

This shows that the resistance of each length of constantan wire is about 50 Ω.

In comparing the brightness of lamps with short copper wires and long constantan wires, we can analyze the voltage drop across the wires due to resistance:

1. Calculate voltage drop across short wires (negligible resistance).
2. Calculate voltage drop across long constantan wires. Using $R_{copper} \approx 0 \, Ω$ and $R_{constantan} = 100 \, Ω$ for two wires:
   • Voltage drop in constantan wires:

$V_{drop} = I \cdot R_{constantan} \approx 0.36 \cdot 100 = 36 \, V$.

Since the total supply voltage is 12 V, the voltage available at the lamps will be significantly lower, resulting in dimmer lamps. Therefore, the demonstration effectively shows that lamps will be dimmer when connected with long constantan wires compared to short copper wires.

Advantage:

One major advantage of using superconductors is their zero electrical resistance. This means that electrical energy can be transmitted over long distances without any energy loss, making the system more efficient.

Difficulty:

However, a significant difficulty is maintaining the superconductors at very low temperatures to keep them in their superconducting state. This requires potentially costly cooling systems and infrastructure, posing challenges in implementation.