In this resistor network, the emf of the supply is 12 V and it has negligible internal resistance - AQA - A-Level Physics - Question 29 - 2017 - Paper 1

To determine the voltmeter reading between points X and Y, we first analyze the resistor network.

1. Identify the Configuration: The resistors are arranged in series and parallel. The 2 Ω and the two 1 Ω resistors are in parallel, while the 3 Ω resistor is connected in series with this combination.

2. Calculate Equivalent Resistance:

   • The equivalent resistance (R_parallel) of the parallel resistors (2 Ω, 1 Ω, and 1 Ω):

     $\frac{1}{R_{parallel}} = \frac{1}{2} + \frac{1}{1} + \frac{1}{1}$

     This simplifies to:

     $R_{parallel} = \frac{1}{\frac{1}{2} + 2} = \frac{1}{2.5} = 0.4 \text{Ω (approximately)}$

   • Total resistance in the circuit (R_total) is then:

     $R_{total} = R_{parallel} + 3 = 0.4 + 3 = 3.4 \text{Ω}$

3. Calculate Current (I): Using Ohm’s law, current through the circuit is:

   $I = \frac{V}{R_{total}} = \frac{12V}{3.4Ω} \approx 3.53A$

4. Voltage Drop Across the Resistors: The voltmeter is connected between points X and Y. Therefore, we need to evaluate the voltage drop:

   • The drop across the 3 Ω resistor:

   $V = I \times R = 3.53A \times 3Ω \approx 10.59V$

   • The remaining voltage across the equivalent parallel resistors:

   $V_{parallel} = 12V - 10.59V \approx 1.41V$

5. Since both identical 1 Ω resistors will have the same voltage drop, the reading on the voltmeter between X and Y will be half of 1.41V, yielding:

   $V_{XY} = \frac{1.41V}{2} \approx 0.705V ≈ 0V ext{ to 1 significant figure.}$

As options only provide whole numbers, the closest option to our calculation is A 0 V.