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.

Cells in Series

When cells are connected in series the total e.m.f. of the series combination is equal to the sum of the e.m.f.’s of the individual cells and the internal resistance of the cells will also be in series.

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.
\(
V_{ AB } \equiv V(A)-V(B)=\varepsilon_1-I r_1
\)
Similarly,
\(
V_{ BC } \equiv V(B)-V( C )=\varepsilon_2-I r_2 \dots(i)
\)
Hence, the potential difference between the terminals A and C of the combination is
\(
\begin{aligned}
V_{ AC } & \equiv V(A)-V( C )=V(A)-V(B)+V(B)-V( C ) \\
& =\left(\varepsilon_1+\varepsilon_2\right)-I\left(r_1+r_2\right) \dots(ii)
\end{aligned}
\)
If we wish to replace the combination by a single cell between \(A\) and C of \(\operatorname{emf} \varepsilon_{e q}\) and internal resistance \(r_{e q}\), we would have
\(
V_{A C}=\varepsilon_{e q}-I r_{e q} \dots(iii)
\)
Comparing the last two equations (ii) and (iii), we get
\(
\varepsilon_{e q}=\varepsilon_1+\varepsilon_2
\)
and \(r_{e q}=r_1+r_2\)
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. (i) would change to \(V_{B C}=-\varepsilon_2-I r_2\) and we will get
\(
\varepsilon_{e q}=\varepsilon_1-\varepsilon_2 \quad\left(\varepsilon_1>\varepsilon_2\right)
\)
The rule for series combination clearly can be extended to any number of cells:

• The equivalent emf of a series combination of \(n\) cells is just the sum of their individual emf’s, and
• The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances.
• Cells in Series Equivalent emf: \(\varepsilon_{e q}=\varepsilon_1+\varepsilon_2+ \dots +\varepsilon_n\) and Equivalent internal resistance \(r_{e q}=r_1+r_2+\ldots+r_n\)

Cells in Series  connected with an external load resistance

If \(n\) identical cells of emf \(E\) and internal resistance \(r\) each, are connected in series, then

• equivalent emf of the combination, \(E_{\text {eq }}=n E\).
• equivalent internal resistance of the combination, \(r_{\text {eq }}=n r\).
• main current \(=\) Current from each cell \(=I=\frac{n E}{R+n r}\).
• potential difference across external resistance, \(V=I R\).
• potential difference across each cell, \(V^{\prime}=\frac{V}{n}\).
• If \(nr \gg R, I =\frac{ E }{ r } \approx\) current from any cell when shot-circuited.
• If \(nr << R, I =\frac{ nE }{ R } \approx n \times\) current from any one cell, when connected with the external resistance.

Example 54: Find the current in the loop. 

Solution: \(E_{\text {eq }}=40-15+20-10=35 V\)
\(r_{\text {eq }}=2+1+1+1=5 \Omega\).
Total resistnace \(= 5 + 10 = 15 \Omega\).
\(
\begin{aligned}
i & =\frac{35}{10+5}=\frac{35}{15} \\
& =\frac{7}{3} A
\end{aligned}
\)

Cells in Parallel

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
\(
I=I_1+I_2 \dots(i)
\)
Let \(V\left(B_1\right)\) and \(V\left(B_2\right)\) be the potentials at \(B_1\) and \(B_2\), respectively. Then, considering the first cell, the potential difference across its terminals is \(V\left(B_1\right)-V\left(B_2\right)\). Hence,
\(
V \equiv V\left(B_1\right)-V\left(B_2\right)=\varepsilon_1-I_1 r_1 \dots(ii)
\)
Points \(B_1\) and \(B_2\) are connected exactly similarly to the second cell. Hence considering the second cell, we also have
\(
V \equiv V\left(B_1\right)-V\left(B_2\right)=\varepsilon_2-I_2 r_2 \dots(iii)
\)
Combining the last three equations
\(
\begin{aligned}
I & =I_1+I_2 \\
& =\frac{\varepsilon_1-V}{r_1}+\frac{\varepsilon_2-V}{r_2}=\left(\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}\right)-V\left(\frac{1}{r_1}+\frac{1}{r_2}\right) \dots(iv)
\end{aligned}
\)
Hence, \(V\) is given by,
\(
V=\frac{\varepsilon_1 r_2+\varepsilon_2 r_1}{r_1+r_2}-I \frac{r_1 r_2}{r_1+r_2} \dots(v)
\)
If we want to replace the combination by a single cell, between \(B_1\) and \(B_2\), of emf \(\varepsilon_{e q}\) and internal resistance \(r_{e q}\), we would have
\(
V=\varepsilon_{e q}-I r_{e q} \dots(vi)
\)
The last two equations should be the same and hence
\(
\begin{aligned}
& \varepsilon_{e q}=\frac{\varepsilon_1 r_2+\varepsilon_2 r_1}{r_1+r_2} \dots(vii) \\
& r_{e q}=\frac{r_1 r_2}{r_1+r_2} \dots(viii)
\end{aligned}
\)
We can put these equations in a simpler way,
\(
\begin{aligned}
& \frac{1}{r_{e q}}=\frac{1}{r_1}+\frac{1}{r_2} \dots(ix) \\
& \frac{\varepsilon_{e q}}{r_{e q}}=\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2} \dots(x)
\end{aligned}
\)

Equations (ix) and (x) can be extended easily. If there are n cells of emf \(\varepsilon_1, \ldots \varepsilon_{ n }\) and of internal resistances \(r_1, \ldots r_{ n }\) respectively, connected in parallel, the combination is equivalent to a single cell of emf \(\varepsilon_{\text {eq }}\) and internal resistance \(r_{e q}\), such that
\(
\begin{aligned}
& \frac{1}{r_{e q}}=\frac{1}{r_1}+\ldots+\frac{1}{r_n} \dots(xi) \\
& \frac{\varepsilon_{e q}}{r_{e q}}=\frac{\varepsilon_1}{r_1}+\ldots+\frac{\varepsilon_n}{r_n} \dots(xii)
\end{aligned}
\)

Cells in Parallel connected with an external load resistance

If \(n\) identical cells of emf \(E\) and internal resistance \(r\) each are connected in parallel, then

• equivalent emf of the combination, \(E_{\text {eq }}=E\).
• equivalent internal resistance, \(R_{\text {eq }}=r / n\).
• main current, \(I=\frac{E}{R+r / n}\).
• current from each cell, \(I^{\prime}=\frac{I}{n}\).
• potential difference across external resistance \(=\) potential difference across each cell \(=V=I R\).
• If \(\text { If } r \ll n R \quad I = E / R =\) Current from any one cell when connected with the external resistance
• If \(r>n R\) \(\quad I =\frac{ nE }{ r }= n \times\) current from any cell when short-circuited. If external resistance is much smaller than the total internal resistance, then cells should be connected in parallel to get the maximum current.

Cells in Multiple Arc

\(mn =\) number of identical cells.
\(n =\) number of rows
\(m =\) number of cells in each row.
The combination of cells is equivalent to single cell of \(emf = mE\)
and internal resistance \(=\frac{ mr }{ n }\)
Current \(I=\frac{m E}{R+\frac{m r}{n}}\)
For maximum current \(n R=m r\) \(R =\frac{ mr }{ n }=\) internal resistance of the equivalent battery.
\(I _{\max }=\frac{ nE }{2 r }=\frac{ mE }{2 R }\).

Mixed grouping

If \(n\) identical cells of emf \(E\) and internal resistance \(r\) each, are connected in a row and such \(m\) rows are connected in parallel as shown below, then

• Equivalent emf of the combination,
  \(
  E_{\text {eq }}=n E
  \)
• Equivalent internal resistance,
  \(
  r_{\text {eq }}=\frac{n r}{m}
  \)
• The main current flowing through the load,
  \(
  I=\frac{n E}{R+\frac{n r}{m}}=\frac{m n E}{m R+n r}
  \)
• The potential difference across the load,
  \(
  V=I R
  \)
• Current from each cell,
  \(
  I^{\prime}=\frac{I}{n}
  \)
• Potential differences across each cell,
  \(
  V^{\prime}=\frac{V}{n}
  \)
• Total number of cells \(=m n\)
  In the mixed grouping of cells, the current through the external resistance would be maximum, if the external resistance is equal to the total internal resistance of the cells, i.e. \(R=\frac{n r}{m}\).

Example 55: A battery of six cells each of e.m.f. 2 V and internal resistance \(0.5 \Omega\) is being charged by D. C. mains of e.m.f. 220 V by using an external resistance of \(10 \Omega\). What will be the charging current?

Solution: Net e.m.f of the battery \(=12 V\) and total internal resistance \(=3 \Omega\)
Total resistance of the circuit \(=3+10=13 \Omega\)
\(
\text { Charging current } I=\frac{\text { Net e.m.f. }}{\text { total resistance }}=\frac{220-12}{13}=16 A
\)

Example 56: A battery of six cells each of e.m.f. 2 V and internal resistance \(0.5 \Omega\) is being charged by D. C. mains of e.m.f. 220 V by using an external resistance of \(10 \Omega\). What is the potential difference across the battery?

Solution: In case of charging of battery, terminal potential \(\quad V = E + Ir =12+16 \times 3=60\) volt.

Example 57: Four identical cells each of e.m.f. 2 V are joined in parallel providing supply of current to external circuit consisting of two \(15 \Omega\) resistors joined in parallel. The terminal voltage of the equivalent cell as read by an ideal voltmeter is 16 V calculate the internal resistance of each cell

Solution:

Total internal resistance of the combination \(r_{\text {eq }}=\frac{r}{4}\)
Total e.m.f. \(E _{\text {eq }}=2 V\)
Total external resistance \(R =\frac{15 \times 15}{15+15}=\frac{15}{2}=7.5 \Omega\)
Current drawn from equivalent cell \(I =\frac{\text { terminal potential }}{\text { external resistance }}=\frac{1.6}{7.5} A\)
\(
\because E-I\left(\frac{r}{4}\right)=1.6 \quad \therefore 2-\frac{1.6}{7.5}\left(\frac{r}{4}\right)=1.6 \Rightarrow r=7.5 \Omega
\)

Example 58: The e.m.f. of a primary cell is 2 V , when it is shorted then it gives a current of 4 A . Calculate internal resistance of primary cell.

Solution: 
\(I=\frac{E}{r+R}\), If cell is shorted then \(R=0, I=\frac{E}{r} \quad \therefore r=\frac{E}{I} \quad=\frac{2}{4}=0.5 \Omega\)

Example 59: \(n\) rows each containing m cells in series, are joined in parallel. Maximum current is taken from this combination in a \(3 \Omega\) resistance. If the total number of cells used is 24 and the internal resistance of each cell is \(0.5 \Omega\), find the value of \(m\) and \(n\).

Solution:

Total number of cell \(mn =24\), For maximum current \(\frac{ mr }{ n }=R \Rightarrow 0.5 m=3 n , m =\frac{3 n }{0.5}=6 n\)
\(
\therefore 6 n \times n =24 \Rightarrow n =2 \text { and } m \times 2=24 \Rightarrow m =12
\)

Example 60: Find the EMF and internal resistance of a single battery which is equivalent to a combination of three batteries as shown in the figure.

Solution:

Battery (B) and (C) are in parallel combination with opposite polarity. So, their equivalent
\(
\varepsilon_{B C}=\frac{\frac{10}{2}+\frac{-4}{2}}{\frac{1}{2}+\frac{1}{2}}=\frac{5-2}{1}=3 V \quad \Rightarrow \quad r_{B C}=1 \Omega .
\)

\(
\varepsilon_{ ABC }=6-3=3 V \quad \Rightarrow \quad r _{ ABC }=2 \Omega \text {. }
\)

Example 61: Two identical cells of emf 1.5 V each joined in parallel provide supply to an external circuit consisting of two resistances of \(17 \Omega\) each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4 V. Calculate the internal resistance of each cell.

Solution: Given, \(E=1.5 V\) and \(V=1.4 V\)
Resistance of external circuit \(=\) Total resistance of two resistances of \(17 \Omega\) connected in parallel
\(
\Rightarrow \quad R=\frac{R_1 R_2}{R_1+R_2}=\frac{17 \times 17}{17+17}=8.5 \Omega
\)

Let \(r^{\prime}\) be the total internal resistance of the two cells.
Then,
\(
r^{\prime}=R\left(\frac{E-V}{V}\right)=8.5\left(\frac{1.5-1.4}{1.4}\right) \Omega=0.6 \Omega
\)
As the two cells of internal resistance \(r\) each have been connected in parallel, therefore
\(
\frac{1}{r^{\prime}}=\frac{1}{r}+\frac{1}{r} \text { or } \frac{1}{0.6}=\frac{2}{r}
\)
or \(r=0.6 \times 2=12 \Omega\)

Example 62: Find the minimum number of cells required to produce an electric current of 1.5 A through a resistance of \(30 \Omega\). Given that the emf of each cell is 1.5 V and internal resistance is \(1.0 \Omega\).

Solution: As, \(\frac{n r}{m}=R\)
\(
\therefore \quad \frac{n \times 1}{m}=30 \text { or } n=30 m \dots(i)
\)
Current, \(I=\frac{n E}{2 R}\) or \(1.5=\frac{n \times 1.5}{2 \times 30}\) or \(n=60 \dots(ii)\)
From Eqs. (i) and (ii), we get
\(
\begin{aligned}
m & =60 / 30=2 \\
m n & =120
\end{aligned}
\)

Example 63: 36 cells, each of internal resistance \(0.5 \Omega\) and emf of 1.5 V, are used to send maximum current through an external circuit of \(2 \Omega\) resistance. Find the best mode of grouping them and the maximum current through the external circuit.

Solution: Given, \(E=1.5 V, r=0.5 \Omega\) and \(R=2 \Omega\)
Total number of cells, \(m n=36 \dots(i)\)
For maximum current in the mixed grouping,
\(
\frac{n r}{m}=R \text { or } \quad \frac{n \times 0.5}{m}=2 \dots(ii)
\)
Multiplying Eqs. (i) and (ii), we get
\(
\begin{aligned}
& 0.5 n^2 =72 \text { or } n^2=144 \\
\therefore & n =12 \text { and } m=\frac{36}{12}=3
\end{aligned}
\)
Thus, for maximum current there should be three rows in parallel, each containing 12 cells in series.
\(
\therefore \text { Maximum current }=\frac{m n E}{m R+n r}=\frac{36 \times 1.5}{3 \times 2+12 \times 0.5}=4.5 A
\)

Example 64: 12 cells, each of emf 1.5 V and internal resistance of \(0.5 \Omega\), are arranged in \(m\) rows each containing \(n\) cells connected in series, as shown. Calculate the values of \(n\) and \(m\) for which this combination would send maximum current through an external resistance of \(1.5 \Omega\).

Solution: For maximum current through the external resistance, external resistance \(=\) total internal resistance of cells
\(
\begin{aligned}
R & =\frac{n r}{m} \\
1.5 & =\frac{n \times 0.5}{\frac{12}{n}} \\
36 & =n^2 \text { or } n=6 \text { and } m=2
\end{aligned} \quad(\because m n=12)
\)