A battery with unknown emf (ε) and unknown internal resistance (r) is connected to three resistors, a high-resistance voltmeter, two switches and two ammeters of negligible resistance, as shown below - NSC Physical Sciences - Question 8 - 2023 - Paper 1

To calculate the voltmeter reading (V), we use the formula:

$V = I imes R_{total}$

First, we need to find the total resistance:

1. The total resistance in the circuit is $R_{total} = R_1 + R_2 + R_3 = 2 \\Omega + 3 \\Omega + 5 \\Omega = 10 \\Omega$

2. So, substituting the values, we get: $V = 1.5 A * 10 \\Omega = 15 V$.

To find the reading on ammeter A2, we need to apply the current division rule. Using Ohm's law with the resistance:

1. The reading on A2 can be calculated as: $I_{A2} = I_{total} imes \frac{R_{3}}{R_{1}+R_{2}+R_{3}}$ $I_{A2} = 1.5 A \times \frac{3}{10} = 0.45 A$.

The power dissipated, P, in a resistor can be calculated using the formula:

$P = I^2 \times R$

Here, we can use the reading obtained for A2: $P = (0.45 A)^2 \times 3 \Ω = 0.2025 A^2 \times 3 \Ω = 0.6075 W$.

With switch S1 OPEN and S2 CLOSED, we can calculate the emf (ε) using the formula:
$ε = I_{A2} \times R_{total} + I_{A2} \times r$ Substituting the current reading:

1. Given: $I_{A2} = 3.64 A$ and the internal resistance can be calculated knowing: $R_{total} = 2 \Ω + 5 \Ω = 7 \Ω$
2. Use the above formula to get: $ε = 3.64 A \times 7 \Ω + 3.64 A \times r$ Thus: Perform backwards substitutions to find ε.

When switch S2 opens with S1 closed, the circuit configuration changes, causing the total resistance to increase which subsequently decreases the current flow. Thus, the voltmeter reading will DECREASE as the voltage drop across the resistors increases.