Two identical cells, each of emf 1.5V and internal resistance 1\(\Omega\), are connected in parallel to supply current to a 2 \(\Omega\) resistor. What is t...

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Two identical cells joined in parallel behave as a single cell whose e.m.f. is the same as one of them, because their terminals are tied together so neither can raise the terminal voltage above its own e.m.f. What the parallel arrangement does change is the internal resistance: the two internal resistances are in parallel, so

\[r_{\text{eff}} = \frac{r}{n} = \frac{1\,\Omega}{2} = 0.5\,\Omega, \qquad E = 1.5\,\mathrm{V}.\]

Applying the circuit equation \(E = I(R + r_{\text{eff}})\) with the external resistor \(R = 2\,\Omega\):

\[I = \frac{E}{R + r_{\text{eff}}} = \frac{1.5}{2 + 0.5} = \frac{1.5}{2.5} = 0.6\,\mathrm{A}.\]

This \(0.6\,\mathrm{A}\) is the total current delivered to the resistor; each cell supplies half of it, \(0.3\,\mathrm{A}\), which is why parallel grouping is used when a circuit needs a larger current than one cell can comfortably provide at the same voltage.

The trap in this question is to treat the cells as though they were in series. That would give \(E = 3.0\,\mathrm{V}\), \(r = 2\,\Omega\) and \(I = 3.0/4 = 0.75\,\mathrm{A}\), which rounds close to one of the other figures offered. A second common slip is to use \(r = 1\,\Omega\) unchanged and obtain \(1.5/3 = 0.5\,\mathrm{A}\). Fix the rule firmly: cells in series add their e.m.f.s and their internal resistances; identical cells in parallel keep the single-cell e.m.f. and divide the internal resistance by the number of cells.