A cell with internal resistance provides a current. Explain why the terminal potential difference (p.d.) is smaller than the electromotive force (e.m.f.) of the cell.
A battery with e.m.f. $12\,\text{V}$ and internal resistance $0.50\,\Omega$ is connected to a variable resistor $X$ and a resistor $Y$ of fixed resistance, as shown in Fig. 7.1. The resistance $R$ of $X$ is raised from $2.0\,\Omega$ to $16\,\Omega$. Figure 7.2 shows how the current $I$ in the circuit varies with $R$. Calculate, for $I = 1.2\,\text{A}$, the p.d. across $X$.
Calculate, for $I = 1.2\,\text{A}$, the resistance of $Y$.
Calculate, for $I = 1.2\,\text{A}$, the power lost in the battery.
Use Fig. 7.2 to explain how the terminal p.d. of the battery changes as the resistance $R$ of $X$ is increased.
Calculate, for $I = 1.2\,\text{A}$, the p.d. across $X$.
Calculate, for $I = 1.2\,\text{A}$, the resistance of $Y$.
Calculate, for $I = 1.2\,\text{A}$, the power dissipated in the battery.