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Physics · Ch 11 — Electric Current Through Conductors

Cells in Parallel

11.13

Cells in Parallel

In a PARALLEL combination of cells, all the POSITIVE terminals of the cells are joined together at one common node, and all the NEGATIVE terminals are joined together at another common node (Fig. 11.13b). Here the total current supplied to the external circuit divides among the individual cells (branches) -- I1I_1 from the first cell, I2I_2 from the second, and so on -- rather than the voltage dividing, as it does in series.

Consider two cells connected this way, with their shared terminals at potentials VB1V_{B1} and VB2V_{B2}. For the FIRST cell (emf ε1\varepsilon_1, internal resistance r1r_1, delivering current I1I_1), the potential difference across its own terminals is

V=VB1−VB2=ε1−I1r1— (11.45)⇒I1=ε1−Vr1— (11.46)V=V_{B1}-V_{B2}=\varepsilon_1-I_1r_1\qquad\text{--- (11.45)}\qquad\Rightarrow\qquad I_1=\frac{\varepsilon_1-V}{r_1}\qquad\text{--- (11.46)}

and, because points B1B_1 and B2B_2 connect to the SECOND cell in exactly the same way, that cell gives the analogous pair

V=ε2−I2r2— (11.47)⇒I2=ε2−Vr2V=\varepsilon_2-I_2r_2\qquad\text{--- (11.47)}\qquad\Rightarrow\qquad I_2=\frac{\varepsilon_2-V}{r_2}

Since the total current is I=I1+I2I=I_1+I_2, substituting the two branch currents from Eq. (11.46) and its analogue for cell 2, and collecting terms, leads (after some algebra) to

V=ε1r2r1+r2+ε2r1r1+r2−Ir1r2r1+r2— (11.49)V=\varepsilon_1\frac{r_2}{r_1+r_2}+\varepsilon_2\frac{r_1}{r_1+r_2}-I\frac{r_1r_2}{r_1+r_2}\qquad\text{--- (11.49)}

If this two-cell combination is now replaced by a SINGLE equivalent cell of emf εeq\varepsilon_{eq} and internal resistance reqr_{eq} connected between the same points B1B_1 and B2B_2,

V=εeq−Ireq— (11.50)V=\varepsilon_{eq}-Ir_{eq}\qquad\text{--- (11.50)}

comparing Eq. (11.49) and Eq. (11.50) term by term gives the equivalent internal resistance and equivalent emf of the parallel pair:

1req=1r1+1r2— (11.51a)εeqreq=ε1r1+ε2r2— (11.51b)\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}\qquad\text{--- (11.51a)}\qquad\qquad \frac{\varepsilon_{eq}}{r_{eq}}=\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}\qquad\text{--- (11.51b)}

These generalise for nn cells (emfs ε1,ε2,…,εn\varepsilon_1,\varepsilon_2,\dots,\varepsilon_n; internal resistances r1,r2,…,rnr_1,r_2,\dots,r_n) connected in parallel to

1req=1r1+1r2+⋯+1rn— (11.52)\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\dots+\frac{1}{r_n}\qquad\text{--- (11.52)}

εeqreq=ε1r1+ε2r2+⋯+εnrn— (11.53)\frac{\varepsilon_{eq}}{r_{eq}}=\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}+\dots+\frac{\varepsilon_n}{r_n}\qquad\text{--- (11.53)}

When substituting emf values into these formulas, each εi\varepsilon_i must be given its correct ALGEBRAIC sign according to the actual polarity with which that particular cell has been connected into the combination. …

Figure 11.13bTwo cells in parallel

What this figure shows. Two cells drawn side by side with their POSITIVE terminals both connected together at a common node (labelled B1) and their NEGATIVE terminals both connected together at another common node (labelled B2), so each cell forms an independent path between the same pair of points B1 and B2. The current supplied by each cell (I1 from the first cell, I2 from the second) is shown flowing out of its own positive terminal toward B1 and combining into the total external current I = I1+I2, with the potential difference V = VB1-VB2 common to both cells' terminals -- the arrangement used to derive the e …