A battery with electromotive force (e.m.f.) $6.0\,\text{V}$ and negligible internal resistance is joined in series with a variable resistor and a uniform resistance wire XY, as shown in Fig. 6.1. Wire XY has length $2.00\,\text{m}$ and resistance $8.0\,\Omega$. The resistance $R$ of the variable resistor is adjusted so that the potential difference across wire XY is $2.4\,\text{V}$.
(a)[2]
Determine the value of $R$.
(b)[2]
Explain why the potential difference $V$ between any two points on wire XY varies in direct proportion to the distance $L$ between those points.
(c(i))[2]
Calculate the value of $E$.
(c(ii))[2]
The value of $R$ is now decreased. State and explain the change that must be made to the position of P on wire XY so that the galvanometer reads zero again.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “correct use of circuit equation $2.4/6.0 = 8/(R+8)$ or an equivalent expression” …