Physics 9702 · AS & A Level · Practical circuits

Practical circuits — practice question

(a)[2]

A resistance wire with uniform cross-sectional area $3.3 \times 10^{-7}\,\text{m}^2$ and length $2.0\,\text{m}$ is made from metal with resistivity $5.0 \times 10^{-7}\,\Omega\,\text{m}$. Show that the wire's resistance is $3.0\,\Omega$.

(b(i))[1]

In the circuit shown in Fig. 6.1, the ends of the resistance wire in (a) are joined to terminals $X$ and $Y$. The cell has an electromotive force (e.m.f.) of $1.50\,\text{V}$ and internal resistance $r$. The potential difference between $X$ and $Y$ is $1.20\,\text{V}$. Calculate the current in the circuit.

(b(ii))[2]

Calculate the internal resistance $r$.

(c)[2]

A galvanometer and a cell with e.m.f. $E$ and negligible internal resistance are connected to the circuit in (b), as shown in Fig. 6.2. The resistance wire between $X$ and $Y$ has a length of $2.0\,\text{m}$. When the connection $P$ is adjusted so that $XP$ is $1.4\,\text{m}$, the galvanometer reading is zero. Determine the e.m.f. $E$ of the cell.

(d)[3]

In Fig. 6.2, the circuit is changed by substituting the original resistance wire with a second resistance wire. The second wire has the same length as the original wire and is made from the same metal. The second wire has a smaller cross-sectional area than the original wire. Connection $P$ is moved along the second wire until the galvanometer reads zero. State and explain whether length $XP$ for the second wire is shorter than, longer than or the same as length $XP$ for the original wire when the galvanometer reading is zero.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Applies $R=\rho L/A$

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