Current Electricity • NEET Revision

Example 3.39

Equivalent EMF (Series + Parallel Cells)

Question

Find the emf and internal resistance of a single battery which is equivalent to a combination of three batteries as shown: a 6 V, 1 Ω battery in series with the parallel combination of a 10 V, 2 Ω battery and a 4 V, 2 Ω battery (like polarities together).

Given

Series battery: \(E_3 = 6\ \text{V},\ r_3 = 1\ \Omega\)
Parallel pair: \(E_1 = 10\ \text{V},\ r_1 = 2\ \Omega\) and \(E_2 = 4\ \text{V},\ r_2 = 2\ \Omega\)

Asked

Equivalent \(E_{eq}\) and \(r_{eq}\) of the whole combination.

Formula

Parallel cells: \(E_p = \dfrac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}},\qquad r_p = \dfrac{r_1 r_2}{r_1 + r_2}\)

Series: emfs add, internal resistances add.

Solution

Parallel pair first: \(E_p = \dfrac{\frac{10}{2} + \frac{4}{2}}{\frac{1}{2} + \frac{1}{2}} = \dfrac{5 + 2}{1} = 7\ \text{V}\).
\(r_p = \dfrac{2 \times 2}{2 + 2} = 1\ \Omega\).
Now add the series battery: \(E_{eq} = 6 + 7 = 13\ \text{V}\).
\(r_{eq} = 1 + 1 = 2\ \Omega\).

\(E_{eq} = 13\ \text{V},\quad r_{eq} = 2\ \Omega\)

Short Cut

For equal internal resistances in parallel, \(E_p\) is just the average of the emfs: \(\frac{10+4}{2} = 7\ \text{V}\), and \(r_p = \frac{r}{2} = 1\ \Omega\). Spot equal \(r\)'s and skip the full Millman formula.