The battery in this circuit has an emf of 4.2 V and negligible internal resistance - AQA - A-Level Physics - Question 27 - 2021 - Paper 1

When the switch is closed, current flows through the circuit. To find the potential difference across the voltmeter, we first need to calculate the total resistance of the circuit. The two 40 Ω resistors are in parallel, so we calculate their equivalent resistance (R_eq):

$R_{eq} = \frac{1}{\frac{1}{40} + \frac{1}{40}} = 20 \: \Omega$

Now, the total resistance in the circuit is:

$R_{total} = 10 + R_{eq} = 10 + 20 = 30 \: \Omega$

According to Ohm's law, we can find the total current (I) flowing through the circuit:

$I = \frac{V}{R_{total}} = \frac{4.2}{30} = 0.14 \: A$

Next, we need to find the voltage across the equivalent resistance (R_eq) using Ohm's law:

$V_{R_{eq}} = I \times R_{eq} = 0.14 \times 20 = 2.8 \: V$

The voltmeter reads the voltage across one of the 40 Ω resistors, which is:

$V_{40} = V_{R_{eq}} = 2.8 \: V$

Thus, the reading on the voltmeter when the switch is closed is approximately 2.8 V.