A battery of an unknown emf and an internal resistance of 0.5 Ω is connected to three resistors, a high-resistance voltmeter and an ammeter of negligible resistance, as shown below - NSC Physical Sciences - Question 9 - 2016 - Paper 1

Using the formula for total current (I_{ ext{total}}) through the circuit:

$I_{ ext{total}} = I + I_{2}$

We know:

• $I = 0.2 \text{ A}$
• The voltage across the 4 Ω resistor can be calculated using the previous answer (1.3 V).
• Using Ohm's law, $I_{2}$ can be calculated as:

$I_{2} = \frac{1.3}{4} = 0.325 \text{ A}$

Thus, total current:

$I_{ ext{total}} = 0.2 + 0.325 = 0.525 \text{ A}$

To find the emf (\varepsilon) of the battery, we can use the formula:

$\varepsilon = I_{total} \times (R_{total} + r)$

Where:

• $r = 0.5 \Omega$ (internal resistance of the battery)
• $R_{total} = 6.5 \Omega$
• So, substituting:

$\varepsilon = 0.525 \times (6.5 + 0.5) = 0.525 \times 7 = 3.675 \text{ V}$

Removing the 2 Ω resistor increases the total resistance of the circuit. This means that the total current decreases, which subsequently increases the voltmeter reading, as the voltage across the resistors in series will increase. Thus:

• The voltmeter reading will INCREASE.
• The explanation is based on Ohm's law: when the total resistance increases, the current decreases, resulting in an increase in voltage observed across the voltmeter.