Current Electricity • NEET Revision

Example 3.40

Cells in Parallel

Question

Two identical cells of emf 1.5 V each, joined in parallel, supply an external circuit consisting of two resistances of 17 Ω each joined in parallel. A very high resistance voltmeter reads the terminal voltage of the cells as 1.4 V. Calculate the internal resistance of each cell.

Given

Two identical cells in parallel: \(E = 1.5\ \text{V}\) each, internal resistance \(r\) each \(\Rightarrow r_{eq} = \dfrac{r}{2}\)
External: two \(17\ \Omega\) in parallel \(\Rightarrow R = 8.5\ \Omega\)
Terminal voltage \(V = 1.4\ \text{V}\)

Asked

Internal resistance \(r\) of each cell.

Formula

\(I = \dfrac{V}{R}\)

\(E - V = I\, r_{eq} = I\,\dfrac{r}{2}\)

Solution

External resistance: \(R = \dfrac{17}{2} = 8.5\ \Omega\); equivalent source: \(E = 1.5\ \text{V},\ r_{eq} = \dfrac{r}{2}\).
Total current: \(I = \dfrac{V}{R} = \dfrac{1.4}{8.5} = 0.1647\ \text{A}\).
Internal drop: \(E - V = 1.5 - 1.4 = 0.1\ \text{V}\).
\(\dfrac{r}{2} = \dfrac{E - V}{I} = \dfrac{0.1}{0.1647} = 0.607\ \Omega\).
\(r = 2 \times 0.607 \approx 1.21\ \Omega\).

\(r \approx 1.21\ \Omega\) per cell

Short Cut

Use the plug-in with the equivalent source: \(r_{eq} = R\big(\tfrac{E}{V} - 1\big) = 8.5\big(\tfrac{1.5}{1.4} - 1\big) = \tfrac{8.5 \times 0.1}{1.4}\), then double it: \(r = \dfrac{2 \times 8.5 \times 0.1}{1.4} = \dfrac{1.7}{1.4} \approx 1.21\ \Omega\).