A battery with an internal resistance of 0,5 Ω and an unknown emf (ε) is connected to three resistors, a high resistance voltmeter and an ammeter of negligible resistance, as shown in the circuit diagram below - NSC Physical Sciences - Question 8 - 2020 - Paper 1

The voltmeter reading decreases because when switch S is closed, the internal resistance of the battery causes a voltage drop. As current flows through the circuit, energy is converted to heat in the internal resistance, which reduces the voltage available across the external resistors.

8.3.1 Reading on the ammeter: The total resistance in the circuit can be calculated as: $R_{total} = R_2 + R_3 = 25 \, \Omega + 15 \, \Omega = 40 \, \Omega$

Using Ohm's law: $I = \frac{V}{R_{total}}$ Thus, $I = \frac{41.64 \, V}{40 \, \Omega} = 1.041 \text{ A}$

8.3.2 Total external resistance of the circuit: The total external resistance was determined above and is: $R_{ext} = 40 \, \Omega$

8.3.3 Emf of the battery: Using the voltage drop across the internal resistance: $ε = V_{meter} + I \cdot r$ Substituting values gives: $ε = 41.64 \, V + (1.041 \, A \cdot 0.5 \, \Omega) = 41.64 \, V + 0.5205 \, V = 42.16 \, V$

This statement is NOT correct. In a series circuit, the current remains constant throughout all components. Therefore, the current through resistor R3 is equal to the current through resistor R2.

The emf of the battery remains the same. The operation of the voltmeter or the current flowing through the circuit does not change the inherent emf of the battery; it remains constant as long as the battery is functioning normally.