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

To find the electromotive force (emf) of the battery, we first need to calculate the total resistance in the circuit. The total circuit resistance includes the internal resistance of the battery (2.5 Ω) and the resistance of the lamp (approximately 9 Ω), giving:

$R_{total} = R_{lamp} + R_{internal} = 9 + 2.5 = 11.5 \Omega$

Using Ohm's Law, we can find the current (I) in the circuit. The terminal pd across the battery is 6.2 V:

$I = \frac{V_{terminal}}{R_{total}} = \frac{6.2}{11.5} \approx 0.539 \, A$

Now, we can calculate the emf using the formula:

$emf = V_{terminal} + I \cdot R_{internal} = 6.2 + (0.539 \times 2.5) \approx 9.01 \, V$

To calculate the resistivity (ρ) of the wire, we can use the formula:

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

Given:

• Resistance (R) = 9 Ω
• Length (L) = 5.0 m
• Diameter (d) = 0.19 mm = 0.19 \times 10^{-3} m
• Area (A) can be calculated using the formula for the area of a circle:

$A = \pi \left( \frac{d}{2} \right)^2$

Substituting the value of d:

$A = \pi \left( \frac{0.19 \times 10^{-3}}{2} \right)^2 \approx 2.827 \times 10^{-8} m^2$

Now substituting these values into the resistivity formula gives:

$\rho = 9 \cdot \frac{2.827 \times 10^{-8}}{5} \approx 5.1 \times 10^{-8} \Omega m$

As the movable contact is moved to increase the length of wire in series with the lamp, the resistance in the circuit increases. Since the total resistance increases while the voltage remains constant, the current flowing through the lamp decreases. As a result, the brightness of the lamp will decrease as the contact is moved to the right.

When the variable resistor is connected in parallel with the lamp, moving the contact to increase the length of wire in parallel effectively reduces the total resistance of the circuit. This leads to an increase in the current flowing through the lamp, consequently increasing its brightness as the contact is moved.