Three resistors with resistances $R_1$, $R_2$ and $R_3$ are arranged as in Fig. 6.1. The current through the resistor network is $I$ and the potential difference across it is $V$. Show that the overall resistance $R$ of the network is given by the equation $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.
A battery with electromotive force (e.m.f.) $6.0\,\text{V}$ and internal resistance $r$ is connected to a resistor of resistance $12\,\Omega$ and a variable resistor $X$, as shown in Fig. 6.2.
Explain, using energy, why the potential difference across the battery’s terminals is less than the e.m.f. of the battery.
A charge of $2.5\,\text{kC}$ passes through the battery. Calculate the total energy transformed by the battery.
Calculate the number of electrons passing through the battery.
The resistance of the two resistors in parallel is $4.8\,\Omega$. Calculate the resistance of $X$.
Use your answer in b(iii) to work out the ratio\n$\dfrac{\text{power dissipated in } X}{\text{power dissipated in } 12\,\Omega \text{ resistor}}$.
The resistance of $X$ is now reduced. Explain why the power output of the battery increases.