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

To calculate the emf ($\text{emf}$) of the battery, we use:

1. First, find the current $I$ using the total resistance: $R_{total} = R_{lamp} + R_{internal} = 9 \Omega + 2.5 \Omega = 11.5 \Omega$

2. Using Ohm's Law: $I = \frac{V}{R_{total}} = \frac{6.2}{11.5} \approx 0.54 A$

3. Now calculate the emf using the equation: $\text{emf} = \text{pd} + I \cdot R_{internal}$ Substituting the known values: $\text{emf} = 6.2 + (0.54)(2.5) \approx 6.2 + 1.35 = 7.55 V$ Hence, the emf of the battery is approximately 7.55 V.

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

• $R = 9.0 \Omega$ (resistance of the wire)
• $L = 5.0 m$ (length of the wire)
• $A = \pi r^2$ (cross-sectional area of the wire)

Using the diameter of 0.19 mm, we can find the radius: $r = \frac{0.19 \times 10^{-3}}{2} = 0.095 \times 10^{-3} m$

Now, calculate the cross-sectional area: $A = \pi (0.095 \times 10^{-3})^2 \approx 2.84 \times 10^{-9} m^2$

Now rearranging the resistivity formula gives: $\rho = R \cdot \frac{A}{L} = 9.0 \cdot \frac{2.84 \times 10^{-9}}{5.0}$ Calculating this: $\rho \approx 5.1 \times 10^{-9} \Omega m$

As the contact is moved to increase the length of the wire in series with the lamp, the total resistance in the circuit increases. According to Ohm's law, this increase in resistance will reduce the current flowing through the lamp. Since brightness is directly related to the current in the lamp, the lamp will become dimmer as the resistance increases.

When the contact is moved to a position where the variable resistor is connected in parallel with the lamp, the total resistance of the circuit decreases. This allows more current to flow through the lamp, resulting in an increase in brightness. The lamp will appear brighter as more of the current bypasses the variable resistor.