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: $\text{emf} = \text{terminal pd} + I \times r$ Where terminal pd is 6.2 V, I is the current, and r is internal resistance. First, compute the current using the lamp's resistance from the previous part: $I = \frac{V}{R} = \frac{6.2}{9} \approx 0.69 A$ Now substituting the values: $\text{emf} = 6.2 + (0.69 \times 2.5) \approx 6.2 + 1.725 = 7.925 V$ Thus, the emf of the battery is approximately 7.93 V.

To find the resistivity ((\rho)) of the wire, we use the formula: $\rho = R \frac{A}{L}$ Where:

• R is resistance (9.0 Ω)
• A is the cross-sectional area, which can be calculated from the diameter (d = 0.19 mm = 0.00019 m): $A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.00019}{2} \right)^2 \approx 2.84 \times 10^{-8} m^2$
• L is the length of the wire (5.0 m). Now substituting the values: $\rho = 9.0 \times \frac{2.84 \times 10^{-8}}{5.0} \approx 5.10 \times 10^{-8} \Omega m$

When the moveable copper contact is moved to increase the length of the wire in series with the lamp, the resistance of the circuit increases. As a result, the total current flowing through the lamp decreases, leading to a dimmer light output. Conversely, if the contact is moved to decrease the wire length, the resistance decreases, allowing more current to flow and brightening the lamp.

By connecting the variable resistor in parallel with the lamp, the overall resistance of the circuit is reduced. When the contact is moved to increase the length of the wire in parallel, the total resistance will change, causing the brightness of the lamp to vary. Initially, the brightness will increase as the total current through the lamp increases due to the lower resistance in the circuit.