Source: High school physics (Chinese)
Problem
A supply of batteries is available, each with EMF $1.5 V$ and internal resistance $1 \Omega$, with a maximum allowable output current of $0.05 A$ per battery. Resistors of various values can be used as series voltage-dividing resistors. A load is to be operated at its rated values of $6 V$ and $0.1 A$.
- What is the minimum number of batteries required? Specify the arrangement and the value of the dividing resistor.
- Calculate the efficiency of the circuit (ratio of useful power to total power).
Minimum $10$ batteries arranged as $5$ in series $\times$ $2$ in parallel. Dividing resistor $R_d = 12.5 \Omega$. Efficiency $\eta = 80\%$.
Since the load requires $0.1 A$ but each battery supplies at most $0.05 A$, at least $m = 2$ parallel branches are needed. Let $n$ be the number of batteries in series per branch.
Combined EMF: $\varepsilon = 1.5 n V$. Combined internal resistance: $r_{\text{int}} = n/m = n/2 \Omega$. Total external current $I = 0.1 A$, so the per-branch current is $0.05 A$ (matching the limit).
Loop equation: $1.5 n = 6 + 0.1 R_d + 0.1 \cdot (n/2)$.
For $n = 4$: $\varepsilon = 6 V$ but $0.1 \cdot 2 = 0.2 V$ is lost in internal resistance, leaving only $5.8 V$ — insufficient.
For $n = 5$: $7.5 = 6 + 0.1 R_d + 0.25$, giving $R_d = 12.5 \Omega$. Total batteries: $5 \times 2 = 10$.
Useful power: $P_u = U_{\text{load}} I = 6 \times 0.1 = 0.6 W$. Total power: $P_{\text{tot}} = \varepsilon I = 7.5 \times 0.1 = 0.75 W$.
Efficiency: $\eta = P_u / P_{\text{tot}} = 0.6 / 0.75 = 80\%$.