Physics 5054 · O Level · Series and parallel circuits

Series and parallel circuits — practice question

A student is studying series and parallel resistor arrangements. The student is given two resistors, X and Y, joined in the circuit illustrated in Fig. 2.1. The resistors are different from each other. The electromotive force (e.m.f.) of the power supply is $3.0\,\text{V}$.
(a(i))[2]

Enter the values shown for $V_X$ on the voltmeter and $I_S$ on the ammeter.

(a(ii))[1]

A resistor’s resistance is found from $\text{resistance} = \frac{\text{voltage across the resistor}}{\text{current in the resistor}}$. Work out $R_X$, the resistance of resistor X.

(a(iii))[1]

Explain why the switch is opened after the potential difference and current readings have been taken.

(a(iv))[1]

The potential difference $V_Y$ across Y is found from $V_Y = 3.0 - V_X$. Use this relation and your value of $V_X$ from (a)(i) to work out $V_Y$. Calculate the resistance $R_Y$ of Y.

(b(i))[2]

Complete the circuit diagram in Fig. 2.3 so that the two resistors are shown in parallel between points W and Z. Draw the voltmeter connected to measure the potential difference $V_P$ across both resistors.

(b(ii))[1]

In theory, the total resistance $R_T$ is given by $R_T = \frac{R_X R_Y}{R_X + R_Y}$. Use this relationship together with your values of $R_X$ and $R_Y$ to find $R_T$.

(b(iii))[2]

The manufacturer states that the combined resistance of resistors X and Y when they are connected in parallel is $2.5\,\Omega$. State whether your value of $R_T$ found in (b)(ii) is the same as the manufacturer’s value. Justify your answer with a calculation. Two quantities may be taken as equal within experimental accuracy if their values differ by no more than $10\%$.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Measured value: $V_X = 0.7\text{ V}$

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