Physics 9702 · AS & A Level · Potential difference and power

Potential difference and power — practice question

A battery with electromotive force (e.m.f.) $E$ and internal resistance $r$ is joined to a variable resistor with resistance $R$, as illustrated in Fig. 6.1. The circuit current is $I$, and the potential difference across the variable resistor is $V$.
(a)[1]

Explain, in terms of energy, why $V$ is smaller than $E$.

(b)[1]

State an equation linking $E$, $I$, $r$ and $V$.

(c)

The resistance $R$ of the variable resistor is changed. The way $V$ varies with $I$ is shown in Fig. 6.2.

(c(i))[1]

Use Fig. 6.2 to show how it can be inferred that the e.m.f. of the battery is $2.8\,\text{V}$.

(c(ii))[2]

Calculate the internal resistance $r$.

(d)

The battery contains $9.2\,\text{kJ}$ of energy. The variable resistor is adjusted so that $V = 2.1\,\text{V}$. Use Fig. 6.2 to:

(d(i))[1]

Calculate resistance $R$.

(d(ii))[1]

Calculate the number of conduction electrons passing through the battery in a time of $1.0\,\text{s}$.

(d(iii))[3]

Determine the time for the energy in the battery to become equal to $1.6\,\text{kJ}$. (Assume that the e.m.f. of the battery and the current in the battery remain constant.)

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Energy is lost as heat in the internal resistance

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