Physics 9702 · AS & A Level · Practical circuits

Practical circuits — practice question

A battery with electromotive force (e.m.f.) $7.8\,\text{V}$ and internal resistance $r$ is linked to a filament lamp, as shown in Fig. 6.1. In $1500\,\text{s}$, a total charge of $750\,\text{C}$ passes through the battery. Over that interval, the filament lamp releases $5.7\,\text{kJ}$ of energy. The e.m.f. of the battery stays constant.
(a(i))[1]

Explain, using energy ideas and without doing any calculation, why the potential difference across the lamp has to be smaller than the e.m.f. of the battery.

(a(ii).1)[2]

Calculate the current in the circuit.

(a(ii).2)[2]

Calculate the potential difference across the lamp.

(a(ii).3)[2]

Calculate the internal resistance of the battery.

(b(i))[1]

A student is given three resistors with resistances $90\,\Omega$, $45\,\Omega$ and $20\,\Omega$. Sketch a circuit diagram to show how two of these three resistors can be connected so that the combined resistance between the terminals shown is $30\,\Omega$. Label the resistance values on your diagram.

(b(ii))[2]

A potential divider circuit is formed by connecting the three resistors to a battery with e.m.f. $9.0\,\text{V}$ and negligible internal resistance. The potential divider circuit gives an output potential difference $V_{\text{OUT}}$ of $3.6\,\text{V}$. The circuit diagram is shown in Fig. 6.2. On Fig. 6.2, label the resistances of all three resistors and the potential difference $V_{\text{OUT}}$.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Energy is dissipated within the internal resistance $r$

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