A wire is $100\,\text{cm}$ long and has a diameter of $0.38\,\text{mm}$. The wire is made of a metal with resistivity $4.5 \times 10^{-7}\,\Omega\,\text{m}$. As shown in Fig. 6.1, the ends $B$ and $D$ of the wire are joined to a cell $X$. Cell $X$ has an electromotive force (e.m.f.) of $2.0\,\text{V}$ and an internal resistance of $1.0\,\Omega$. A cell $Y$ with e.m.f. $1.5\,\text{V}$ and internal resistance $0.50\,\Omega$ is connected to the wire between points $B$ and $C$, as shown in Fig. 6.1. Point $C$ is a distance $l$ from point $B$. The current in cell $Y$ is zero.
(a)[3]
Show that the wire’s resistance is $4.0\,\Omega$.
(b(i))[2]
Calculate the current in cell $X$.
(b(ii))[1]
Calculate the potential difference (p.d.) across wire $BD$.
(b(iii))[2]
Calculate the distance $l$.
(c)[2]
The connection at $C$ is shifted so that $l$ is increased. Explain why the e.m.f. of cell $Y$ is less than its terminal p.d.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “$R = \rho l/A$” …