A battery of emf 12 V and internal resistance 0.2 Ω is connected to three resistors, a high-resistance voltmeter and two switches, an ammeter and connecting wires of negligible resistance, as shown in the circuit diagram below - NSC Physical Sciences - Question 8 - 2024 - Paper 1

The total current, I, when both switches are closed is given by: $I = 5.5 \, A$. Using Ohm's law, the total voltage drop across the circuit can be expressed as: $ext{emf} = I(R_{internal} + R_X + R_Y)$ Substituting known values: $12 = 5.5(0.2 + R_X + 2R_X)$ Solving for $R_X$ gives: $12 = 5.5(0.2 + 3R_X)$ $12 = 1.1 + 16.5R_X$ $10.9 = 16.5R_X$ $R_X = 0.66 \, ext{Ω}$ Then: $R_Y = 2R_X = 2 imes 0.66 = 1.32 \, ext{Ω}$

From the earlier calculation, we have: $R_X = 0.66 \text{Ω}$. The current through RX when both switches are closed is still 5.5 A. The power dissipated in R_X is given by: $P = I^2 R_X = (5.5)^2 imes 0.66 \approx 20.07 \, W$.

When both switches are opened, the reading on the ammeter is 1.3 A. The voltage across RY can be calculated using Ohm's law: $V = I imes R_Y = 1.3 \times 1.32 \approx 1.72 \, V$. The voltmeter reading will match this voltage drop across RY.

When switch S1 is open and S2 is closed, the total voltage must be reconsidered. The emf across RY when only S2 is closed can be calculated: Using $R_Y = 1.32 \text{Ω}$, $I = \frac{12 \, V}{R_Y + R_{internal}} = \frac{12}{1.32 + 0.2} \approx 8.58 A.$ Thus the new reading on the ammeter will be approximately 8.58 A.