In Fig. 5.1, a $12\,\text{V}$ power supply with negligible internal resistance is connected to a uniform metal wire AB. This wire is $1.00\,\text{m}$ long and has resistance $10\,\Omega$. A $4.0\,\Omega$ resistor and a $2.0\,\Omega$ resistor are joined in series across the wire. The currents $I_1$, $I_2$ and $I_3$ are labelled as in Fig. 5.1.
(a(i))[1]
Use Kirchhoff’s first law to give a relationship linking $I_1$, $I_2$ and $I_3$.
(a(ii))[3]
Calculate the value of $I_1$.
(a(iii))[3]
Calculate the ratio $x$, where $x = \frac{\text{power in metal wire}}{\text{power in series resistors}}$.
(b)[3]
Calculate the potential difference (p.d.) between points $C$ and $D$, as shown in Fig. 5.1. The distance $AC$ is $40\,\text{cm}$ and $D$ lies between the two series resistors.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “So, $I_1 = I_2 + I_3$.” …