A battery has an e.m.f - Edexcel - A-Level Physics - Question 2 - 2023 - Paper 3

To calculate the terminal potential difference (p.d.) of the battery when bulb 1 is lit, we first need to find the resistance of bulb 1 when it operates at its normal working power.

1. Calculate the Resistance of the Bulb: We know that the power (P) of the bulb is given by the formula:

   $P = \frac{V^2}{R}$

   Rearranging for resistance (R), we get:

   $R = \frac{V^2}{P}$

   Where the voltage (V) across the bulb is 12 V, and the power (P) is 40 W:

   $R = \frac{(12)^2}{40} = \frac{144}{40} = 3.6 \Omega$

2. Calculate the Current (I): Since bulb 1 is lit, the current through the circuit can be determined using Ohm’s Law. The total resistance in the circuit when the switch is open is the internal resistance of the battery plus the resistance of bulb 1:

   Total Resistance = Internal Resistance + Bulb Resistance = 0.5 Ω + 3.6 Ω = 4.1 Ω.

   Using the battery’s e.m.f. (E = 12 V):

   $I = \frac{E}{R_{total}} = \frac{12 V}{4.1 \Omega} \approx 2.93 A$

3. Calculate the Terminal P.d.: To find the terminal p.d. when bulb 1 is lit, we can use the formula:

   $V_{terminal} = E - I \times r$

   Substituting in the known values:

   $V_{terminal} = 12 V - (2.93 A \times 0.5 \Omega)$ $= 12 V - 1.465 V \approx 10.5 V$

Thus, the terminal p.d. of the battery when bulb 1 is lit is approximately 10.5 V.