Question

How will you connect (series and parallel) 24 cells each of internal resistance 1 $\Omega$ to get maximum power output across a load of 10 $\Omega$?

Answer 1

Connect 12 cells in series to form each row, and connect 2 such rows in parallel.

Answer 2

For maximum power transfer, the total internal resistance of the battery combination ($r_{total}$) should be equal to the load resistance ($R$). Let $m$ be the number of cells in series in each row, and $n$ be the number of parallel rows. The total number of cells is $N = m \times n = 24$. The total internal resistance of the combination is $r_{total} = \frac{m \times r}{n}$, where $r$ is the internal resistance of a single cell.

Given $r = 1 \Omega$ and $R = 10 \Omega$. For maximum power transfer, $r_{total} = R$, so $\frac{m \times 1}{n} = 10$, which means $m = 10n$.

We have two equations:

1. $m \times n = 24$
2. $m = 10n$

Substituting (2) into (1): $(10n) \times n = 24$ $10n^2 = 24$ $n^2 = 2.4$ $n = \sqrt{2.4}$

Since $m$ and $n$ must be integers, we cannot achieve perfect matching. We need to find integer pairs $(m, n)$ such that $m \times n = 24$ and $\frac{m}{n}$ is closest to $10$.

Let's list possible integer pairs $(m, n)$ and their corresponding $r_{total} = \frac{m}{n}$:

• $(m=1, n=24) \implies r_{total} = 1/24 \approx 0.04 \Omega$
• $(m=2, n=12) \implies r_{total} = 2/12 \approx 0.167 \Omega$
• $(m=3, n=8) \implies r_{total} = 3/8 = 0.375 \Omega$
• $(m=4, n=6) \implies r_{total} = 4/6 \approx 0.667 \Omega$
• $(m=6, n=4) \implies r_{total} = 6/4 = 1.5 \Omega$
• $(m=8, n=3) \implies r_{total} = 8/3 \approx 2.667 \Omega$
• $(m=12, n=2) \implies r_{total} = 12/2 = 6 \Omega$
• $(m=24, n=1) \implies r_{total} = 24/1 = 24 \Omega$

The ideal $r_{total}$ is $10 \Omega$. The closest values are $6 \Omega$ and $24 \Omega$. The power delivered to the load is $P = \frac{m^2 \mathcal{E}^2 R}{(R + r_{total})^2}$, where $\mathcal{E}$ is the emf of a single cell. To maximize power, we need to maximize the term $\frac{R}{(R + r_{total})^2}$ for a fixed $m$ and $\mathcal{E}$. Alternatively, considering the total emf $E_{total} = m\mathcal{E}$ and total internal resistance $r_{total} = \frac{mr}{n}$, the power is $P = \frac{E_{total}^2 R}{(R + r_{total})^2}$. The power delivered is also given by $P = \frac{N \cdot r_{total} \cdot \mathcal{E}^2 \cdot R}{(R + r_{total})^2}$ where $N=24$. $P = \frac{24 \cdot r_{total} \cdot \mathcal{E}^2 \cdot 10}{(10 + r_{total})^2}$. We need to maximize $f(r_{total}) = \frac{r_{total}}{(10+r_{total})^2}$. The derivative $f'(r_{total}) = \frac{10-r_{total}}{(10+r_{total})^3}$. Setting $f'(r_{total})=0$ gives $r_{total}=10$.

Let's evaluate $f(r_{total})$ for the closest values:

• For $r_{total} = 6 \Omega$ (when $m=12, n=2$): $f(6) = \frac{6}{(10+6)^2} = \frac{6}{256}$.
• For $r_{total} = 24 \Omega$ (when $m=24, n=1$): $f(24) = \frac{24}{(10+24)^2} = \frac{24}{1156}$.

Comparing the values: $\frac{6}{256} \approx 0.0234375$ $\frac{24}{1156} \approx 0.02076$

Since $\frac{6}{256} > \frac{24}{1156}$, the power output is maximized when $r_{total} = 6 \Omega$. This corresponds to the configuration where $m=12$ (cells in series per row) and $n=2$ (number of parallel rows). Therefore, connect 12 cells in series to form each row, and connect 2 such rows in parallel.