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A student assembles the circuit in Figure 6 - AQA - A-Level Physics - Question 4 - 2022 - Paper 1
Question 4
A student assembles the circuit in Figure 6. The battery has an internal resistance of 2.5 Ω. Show that the resistance of the 6.2 V, 4.5 W lamp at its working pote... show full transcript
Worked Solution & Example Answer:A student assembles the circuit in Figure 6 - AQA - A-Level Physics - Question 4 - 2022 - Paper 1
Step 1
Show that the resistance of the 6.2 V, 4.5 W lamp at its working potential difference (pd) is about 9 Ω.
Answer
To find the resistance of the lamp, we can use the formula for power: P = rac{V^2}{R} Rearranging this gives: R = rac{V^2}{P} Substituting the values: R = rac{(6.2)^2}{4.5} \\ = rac{38.44}{4.5} \\ = 8.55 \ ext{Ω} Thus, the resistance is approximately 9 Ω.
Step 2
Calculate the emf of the battery.
Answer
Given that the terminal pd across the battery is 6.2 V and the internal resistance of the battery is 2.5 Ω, we can first calculate the total current in the circuit. The resistance of the lamp is about 9 Ω, making the total circuit resistance: Using Ohm's Law, we calculate the current: Now we can calculate the emf of the battery using:
= 6.2 + (0.54 \times 2.5) \\ = 6.2 + 1.35 = 7.55 \text{ V}$$Step 3
Calculate the resistivity of the wire.
Answer
The resistivity can be calculated using the formula: Here, the resistance (R) is 9.0 Ω, the length (L) is 5.0 m, and the area (A) can be calculated from the diameter of 0.19 mm. First, convert the diameter to meters: The radius (r) is: So, the area is:
= \pi (0.000009025) \approx 2.83 \times 10^{-5} \text{ m}^2$$ Substituting the values in: $$\rho = 9.0 \times \frac{2.83 \times 10^{-5}}{5.0} \approx 5.1 \times 10^{-5} \Omega \, ext{m}$$Step 4
Explain, without calculation, what happens to the brightness of the lamp as the contact is moved.
Answer
As the contact is moved to increase the length of the wire in series with the lamp, the resistance of the overall circuit increases. This increase in resistance leads to a decrease in the current flowing through the lamp. Consequently, the brightness of the lamp dims because less current means less power being consumed by the lamp.
Step 5
The contact is returned to its original position on the tube as shown in Figure 8 and the lamp is dim. The contact is again slowly moved to the right as the two plugs are connected.
Answer
When the variable resistor is connected in parallel with the lamp and the contact is moved, it reduces the overall resistance in the circuit. This decrease in resistance allows more current to flow through the lamp than in the previous configuration. Therefore, the brightness of the lamp will increase as more current results in greater power consumption, making the lamp brighter.