Physics · Ch 2 — Current Electricity

Cells in Parallel

2.4.4

Cells in Parallel

Alternatively, n identical cells can be connected in parallel: all of their positive terminals are joined together at one common node, and all of their negative terminals are joined together at a second common node, with these two nodes then forming the terminals of the combined battery, across which an external resistance R is connected (Figure 2.22).

Because the cells are in parallel, the combined battery's emf stays equal to that of just ONE cell, $\varepsilon_{eq}=\varepsilon$ (not multiplied by n, unlike the series case) -- but the n internal resistances, each r, now combine as resistors in parallel would, giving a much SMALLER equivalent internal resistance, $r_{eq}=r/n$ . The total current supplied to R is therefore

$I = \dfrac{\varepsilon}{R+r/n} \qquad (2.42\text{-analogue})$ …

Figure 2.22Cells in parallel

What this figure shows. n cells, each of emf $\varepsilon$ , are drawn with all of their positive terminals joined together at one common node and all of their negative terminals joined together at a second common node; an external resistance R is connected between these two common nodes, and the combined current I from the whole parall …

Misc Example 2.19Parallel battery of four cells driving an external resistor

Worked out. Four cells, each 5 V with internal resistance $0.5\ \Omega$ , are connected in parallel with an external resistance $R=10\ \Omega$ ; several quantities are required. (i) Equivalent emf $\varepsilon_{eq}=\varepsilon=5$ V (parallel wiring does not add up the emf). (ii) Equivalent internal resistance $R_{eq}=r/n=0.5/4=0.125\ \Omega$ . (iii) Total current $I=\varepsilon/(R+r/n)=5/(10+0.125)\approx0.5$ A. (iv) Potential difference across each cell equals the terminal voltage $V=IR=0.5\times10=5$ V. (v) Current supplied by each individual cell is $ …

Start your 14-day free trial to unlock the full solution →

← Previous · 2.4.3Cells in Series Next · 2.5 →Kirchhoff's Rules