Problem — two cells in parallel

Problem

Two cells in parallel. You connect two identical alkaline cells, both nominally $1{,}5\ \text{V}$, in parallel ($+$ terminal to $+$ terminal, $-$ terminal to $-$ terminal), and then attach the whole thing to a light bulb. (a) Does the bulb shine brighter, dimmer, or the same compared to when it is powered by a single cell? (b) How long does the parallel configuration last, compared to a single cell (for the same bulb)? (c) If one of the two cells is nearly exhausted (EMF $\sim 1{,}2\ \text{V}$), what happens as soon as you close the circuit? In which direction does the current flow in the “good” cell? Answer by reasoning about internal resistance.

Solution

(a) SAME brightness. Two identical cells in parallel maintain the same EMF $\mathcal{E} = 1{,}5\ \text{V}$ at the common terminals: the voltage applied to the bulb does not change, so the current $i = \mathcal{E}/R$ and the power $\mathcal{E}^2/R$ are the same as in the single-cell case. (In addition, the two internal resistances go in parallel, $r/2$, slightly reducing the internal drop: a negligible effect with a bulb of much greater resistance.)

(b) ROUGHLY DOUBLE duration. In parallel, each cell supplies only half the bulb’s current. The total charge available is the sum of that of the two cells: for the same current drawn by the load, the configuration lasts roughly twice as long as a single cell.

(c) Rebalancing current in the exhausted cell. As soon as you close the circuit, the two cells have different EMFs ($1{,}5\ \text{V}$ and $1{,}2\ \text{V}$) but are connected to the same terminals. The difference $\mathcal{E}_1 - \mathcal{E}_2 = 0{,}3\ \text{V}$ drives a current through the internal resistances of the two cells alone: the good cell ($1{,}5\ \text{V}$) supplies current that partly feeds the bulb and partly enters the flat cell, partially “recharging” it. In the good cell the current leaves from the $+$ terminal (normal discharge); in the exhausted cell the current enters through the $+$ terminal (opposite to the discharge direction). The internal resistance limits this rebalancing current.

$\ev{\text{(a) same}\;;\quad \text{(b) duration} \approx \text{double}\;;\quad \text{(c) the good cell "recharges" the flat one: current flows from the good to the exhausted cell}}$

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Topics: Electric circuits Concepts: Electromotive force