3.12 CELLS IN SERIES AND IN PARALLEL

Like resistors, cells can be combined together in an electric circuit. And like resistors, one can, for calculating currents and voltages in a circuit, replace a combination of cells by an equivalent cell.

Consider first two cells in series (Fig. 3.20), where one terminal of the two cells is joined together leaving the other terminal in either cell free. \( \varepsilon_1 \) , \( \varepsilon_2 \) are the emf’s of the two cells and \( r_1 \), \( r_2 \) their internal resistances, respectively.

Let \( V (A) \), \( V (B) \), \( V (C) \) be the potentials at points \( A \), \( B \) and \( C \) shown in Fig. 3.20. Then \( V (A) – V (B) \) is the potential difference between the positive and negative terminals of the first cell. We have already calculated it in Eq. (3.57) and hence,

\[ V_{\mathrm{AB}} \equiv V(\mathrm{~A})-V(\mathrm{~B})=\varepsilon_{1}-I r_{1} \](3.60)

Similarly,

\[ V_{\mathrm{BC}} \equiv V(\mathrm{~B})-V(\mathrm{C})=\varepsilon_{2}-I r_{2} \](3.61)

Hence, the potential difference between the terminals \( A \) and \( C \) of the combination is

\[ \begin{aligned} V_{\mathrm{AC}} & \equiv V(\mathrm{~A})-V(\mathrm{C})\\ &=V(\mathrm{~A})-V(\mathrm{~B})+V(\mathrm{~B})-V(\mathrm{C}) \\ &=\left(\varepsilon_{1}+\varepsilon_{2}\right)-I\left(r_{1}+r_{2}\right) \end{aligned} \](3.62)

If we wish to replace the combination by a single cell between \( A \) and \( C \) of emf \( \varepsilon_{eq} \) and internal resistance \( r_{eq} \), we would have

\[ V_{A C}=\varepsilon_{e q}-I r_{e q} \](3.63)

Comparing the last two equations, we get

\[ \varepsilon_{e q}=\varepsilon_{1}+\varepsilon_{2} \](3.64)

and

\[ r_{e q}=r_{1}+r_{2} \](3.65)

In Fig.3.20, we had connected the negative electrode of the first to the positive electrode of the second. If instead we connect the two negatives, Eq. (3.61) would change to \( V_{BC}=-\varepsilon_2-Ir_2 \) and we will get

\[ \varepsilon_{e q}=\varepsilon_{1}-\varepsilon_{2} \quad\left(\varepsilon_{1}>\varepsilon_{2}\right) \](3.66)

The rule for series combination clearly can be extended to any number of cells:
(i) The equivalent emf of a series combination of n cells is just the sum of their individual emf’s, and
(ii) The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances.

This is so, when the current leaves each cell from the positive electrode. If in the combination, the current leaves any cell from the negative electrode, the emf of the cell enters the expression for \( \varepsilon_{eq} \) with a negative sign, as in Eq. (3.66).

Next, consider a parallel combination of the cells (Fig. 3.21). \( I_1 \) and \( I_2 \) are the currents leaving the positive electrodes of the cells. At the point \( B_1 \), \( I_1 \) and \( I_2 \) flow in whereas the current \( I \) flows out. Since as much charge flows in as out, we have

\[ I=I_1+I_2 \](3.67)

Let \( V (B_1) \) and \( V (B_2) \) be the potentials at \( B_1 \) and \( B_2 \), respectively. Then, considering the first cell, the potential difference across its terminals is \( V (B_1) – V (B_2) \). Hence, from Eq. (3.57)

\[ V \equiv V\left(B_{1}\right)-V\left(B_{2}\right)=\varepsilon_{1}-I_{1} r_{1} \](3.68)

Points \( B_1 \) and \( B_2 \) are connected exactly similarly to the second cell. Hence considering the second cell, we also have

\[ V \equiv V\left(B_{1}\right)-V\left(B_{2}\right)=\varepsilon_{2}-I_{2} r_{2} \](3.69)

Combining the last three equations

\[ \begin{aligned} I &=I_{1}+I_{2} \\ &=\frac{\varepsilon_{1}-V}{r_{1}}+\frac{\varepsilon_{2}-V}{r_{2}} \\ &=\left(\frac{\varepsilon_{1}}{r_{1}}+\frac{\varepsilon_{2}}{r_{2}}\right)-V\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}\right) \end{aligned} \](3.70)

Hence, \( V \) is given by,

\[ V=\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}-I \frac{r_{1} r_{2}}{r_{1}+r_{2}} \](3.71)

If we want to replace the combination by a single cell, between \( B_1 \) and \( B_2 \), of emf \( \varepsilon_{eq} \) and internal resistance \( r_{eq} \), we would have

\[ V=\varepsilon_{e q}-I r_{e q} \](3.72)

The last two equations should be the same and hence

\[ \varepsilon_{e q}=\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} \Gamma_{1}}{r_{1}+r_{2}} \](3.73)

\[ r_{e q}=\frac{r_{1} r_{2}}{r_{1}+r_{2}} \](3.74)

We can put these equations in a simpler way,

\[ \frac{1}{r_{e q}}=\frac{1}{r_{1}}+\frac{1}{r_{2}} \](3.75)

\[ \frac{\varepsilon_{e q}}{r_{e q}}=\frac{\varepsilon_{1}}{r_{1}}+\frac{\varepsilon_{2}}{r_{2}} \](3.76)

In Fig. (3.21), we had joined the positive terminals together and similarly the two negative ones, so that the currents \( I_1 \), \( I_2 \) flow out of positive terminals. If the negative terminal of the second is connected to positive terminal of the first, Eqs. (3.75) and (3.76) would still be valid with \[ \varepsilon_{2} \rightarrow-\varepsilon_{2} \]

Equations (3.75) and (3.76) can be extended easily. If there are \( n \) cells of emf \( \varepsilon_1 \) , . . . \( \varepsilon_n \) and of internal resistances \( r_1 \),... \( r_n \) respectively, connected in parallel, the combination is equivalent to a single cell of emf \( \varepsilon_{eq} \) and internal resistance \( r_{eq} \), such that

\[ \frac{1}{r_{e q}}=\frac{1}{r_{1}}+\ldots+\frac{1}{r_{n}} \](3.77)

\[ \frac{\varepsilon_{e q}}{r_{e q}}=\frac{\varepsilon_{1}}{r_{1}}+\ldots+\frac{\varepsilon_{n}}{r_{n}} \](3.78)