A student assembles the circuit in Figure 6 - AQA - A-Level Physics - Question 4 - 2022 - Paper 1

To calculate the emf of the battery, we use the formula:

$\text{emf} = \text{terminal pd} + I \cdot r$

Where:

• Terminal pd = 6.2 V
• Internal resistance (r) = 2.5 Ω
• Current (I) can be deduced from the resistance of the lamp. Using Ohm's law:

$I = \frac{V}{R} = \frac{6.2}{9} \approx 0.689 \, A$

Substituting the values:

$\text{emf} = 6.2 + (0.689 \times 2.5) \approx 6.2 + 1.723 \approx 7.923 \, V$

Thus, the emf of the battery is approximately 7.92 V.

The resistivity ((\rho)) can be calculated using the formula:

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

• (R = 9.0 , \Omega)
• (L = 5.0 , m)
• (A) is the cross-sectional area of the wire.

The cross-sectional area (A) is given by:

$A = \pi r^2$

First, calculate the radius:

• Diameter = 0.19 mm = 0.19 \times 10^{-3} m, thus radius (r = \frac{0.19}{2} \times 10^{-3} = 0.095 \times 10^{-3} m)

Now calculate the area: $A = \pi (0.095 \times 10^{-3})^2 \approx 2.835 \times 10^{-9} \, m^2$

Substituting the values into the resistivity formula: $9.0 = \rho \frac{5.0}{2.835 \times 10^{-9}}$

Solving for (\rho): $\rho = 9.0 \times \frac{2.835 \times 10^{-9}}{5.0} \approx 5.08 \times 10^{-9} \, \Omega m$

Thus, the resistivity of the wire is approximately (5.08 \times 10^{-9} , \Omega m).

As the movable copper contact on the variable resistor is moved to increase the length of the wire in series with the lamp, the resistance of the circuit increases. Consequently, the current flowing through the lamp decreases. Since the brightness of the lamp is directly proportional to the current through it, the lamp will appear dimmer as the contact is moved. This demonstrates the functionality of the variable resistor in controlling the brightness of the lamp.

When the contact is moved back to its original position, the resistance in series with the lamp is reduced, allowing more current to flow through the lamp. As a result, the brightness of the lamp increases, making it brighter than it was in the dim state. However, if the contact is moved to the right again, the resistance increases once more, leading to reduced current flow and decreasing the brightness of the lamp. This illustrates the relationship between resistance, current, and brightness in an electric circuit.