Physics · Ch 2 — Current Electricity

Cells in Series

2.4.3

Cells in Series

Several identical cells can be strung together in series: the negative terminal of one cell is joined to the positive terminal of the next, and so on down the chain, so that the free positive terminal of the very first cell and the free negative terminal of the very last cell become the two terminals of the resulting battery. For n cells, each of emf $\varepsilon$ and internal resistance r, connected in series with an external resistance R (Figure 2.21):

The n individual internal resistances themselves add up exactly as resistors in series would, giving an equivalent internal resistance $r_{eq}=nr$ ; the total emf of the combination is simply the sum $n\varepsilon$ (the potential difference between the battery's two free terminals). The current in the circuit follows directly from Ohm's law applied to the whole loop:

$I = \dfrac{n\varepsilon}{nr+R} \qquad (2.39)$ …

Figure 2.21Cells in series

What this figure shows. n cells, each of emf $\varepsilon$ shown as identical battery symbols, are drawn chained end to end (negative terminal of one joined to positive terminal of the next), with the free positive terminal of the first cell and free negative terminal of the last cell connected through an external resistance R, and the single current I …

Misc Example 2.18Series battery of four cells driving an external resistor

Worked out. Four cells, each 9 V with internal resistance $0.1\ \Omega$ , are connected in series with an external resistance $R=10\ \Omega$ ; several quantities are required. (i) Equivalent emf $\varepsilon_{eq}=n\varepsilon=4\times9=36$ V. (ii) Equivalent internal resistance $r_{eq}=nr=4\times0.1=0.4\ \Omega$ . (iii) Total current $I=n\varepsilon/(nr+R)=36/(10+0.4)=36/10.4\approx3.46$ A. (iv) Potential difference across R is $V=IR=3.46\times10=34.6$ V, with the remaining $36-34.6=1.4$ V dropped across the internal resistances of the four cells combined. (v) Potential difference across each individua …

Misc ~note-car-headlightsWhy headlights dim when a car engine starts

Worked out. A short practical note: when a car's starter motor draws a very large current from the battery to crank the engine, the headlights sometimes visibly dim at that same moment -- a direct, everyday demonstration of the battery's own internal resistance, since a larger drawn current means a larger voltage drop $Ir$ across the internal resistance, leaving less terminal voltage available for the headlight cir …

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

← Previous · 2.4.2Determination of Internal Resistance Next · 2.4.4 →Cells in Parallel