Cells in Series and in Parallel

Cells in series — the derivation

• Two cells of emf $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ are joined so that the negative terminal of the first touches the positive terminal of the second. Label the free end of the first A, the junction B and the free end of the second C, and let $I$ be the current.
• For the first cell, $V_{AB} = V(\text{A}) - V(\text{B}) = \varepsilon_1 - Ir_1$.
• For the second cell, $V_{BC} = V(\text{B}) - V(\text{C}) = \varepsilon_2 - Ir_2$.
• Adding, the potential difference across the whole combination is

$$V_{AC} = V_{AB} + V_{BC} = (\varepsilon_1 + \varepsilon_2) - I(r_1 + r_2)$$

• Replacing the pair by a single cell between A and C would give $V_{AC} = \varepsilon_{eq} - Ir_{eq}$. Comparing,

$$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2, \qquad r_{eq} = r_1 + r_2$$

• Extended to $n$ cells: the equivalent emf is the sum of the individual emfs and the equivalent internal resistance is the sum of the individual internal resistances — provided the current leaves each cell from its positive electrode.
• If instead the two negative terminals are joined, $V_{BC} = -\varepsilon_2 - Ir_2$ and

$$\varepsilon_{eq} = \varepsilon_1 - \varepsilon_2 \qquad (\varepsilon_1 > \varepsilon_2)$$

• In general, a cell whose current leaves from its negative electrode enters $\varepsilon_{eq}$ with a minus sign. Its internal resistance always adds.
• $n$ identical cells of emf $\varepsilon$ and internal resistance $r$ in series across $R$: $I = n\varepsilon/(R + nr)$. This is the useful arrangement when $R \gg nr$.

Cells in parallel — the derivation

• Two cells are connected with their positive terminals joined at $B_1$ and their negative terminals joined at $B_2$. Let $I_1$ and $I_2$ leave the positive electrodes; by the junction rule the total current is

$$I = I_1 + I_2$$

• Let $V = V(B_1) - V(B_2)$ be the common terminal potential difference. Considering each cell in turn,

$$V = \varepsilon_1 - I_1r_1, \qquad V = \varepsilon_2 - I_2r_2$$

$$\Rightarrow \quad I_1 = \dfrac{\varepsilon_1 - V}{r_1}, \qquad I_2 = \dfrac{\varepsilon_2 - V}{r_2}$$

• Adding,

$$I = \dfrac{\varepsilon_1 - V}{r_1} + \dfrac{\varepsilon_2 - V}{r_2} = \left(\dfrac{\varepsilon_1}{r_1} + \dfrac{\varepsilon_2}{r_2}\right) - V\left(\dfrac{1}{r_1} + \dfrac{1}{r_2}\right)$$

• Solving for $V$,

$$V = \dfrac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} - I\left(\dfrac{r_1 r_2}{r_1 + r_2}\right)$$

• Comparing with $V = \varepsilon_{eq} - Ir_{eq}$,

$$\varepsilon_{eq} = \dfrac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}, \qquad r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}$$

• These are more memorably written, and extend immediately to $n$ cells, as

$$\dfrac{1}{r_{eq}} = \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dots + \dfrac{1}{r_n}, \qquad \dfrac{\varepsilon_{eq}}{r_{eq}} = \dfrac{\varepsilon_1}{r_1} + \dfrac{\varepsilon_2}{r_2} + \dots + \dfrac{\varepsilon_n}{r_n}$$

• If the negative terminal of the second cell is joined to the positive terminal of the first, the same equations hold with $\varepsilon_2 \to -\varepsilon_2$.
• $n$ identical cells of emf $\varepsilon$ and internal resistance $r$ in parallel across $R$: $\varepsilon_{eq} = \varepsilon$, $r_{eq} = r/n$, and $I = \varepsilon/(R + r/n)$. This is the useful arrangement when $R \ll r$.

Diagram in words

• Series: cells drawn one after another along a line, long bar of one facing the short bar of the next, so the arrows of emf all point the same way round the loop; the external resistance $R$ closes the loop from the last free terminal back to the first.
• Parallel: both positive plates wired to one node $B_1$, both negative plates to a second node $B_2$, and $R$ connected between $B_1$ and $B_2$. The branch currents $I_1$ and $I_2$ leave the positive plates and merge into $I$ at $B_1$.

Comparison

Worked cases CBSE has set

• Two cells of 1.5 V, $0.2\ \Omega$ and 2.0 V, $0.3\ \Omega$ in parallel:

$$\varepsilon_{eq} = \dfrac{1.5\times0.3 + 2.0\times0.2}{0.5} = \dfrac{0.85}{0.5} = 1.7\ \text{V}, \qquad r_{eq} = \dfrac{0.2\times0.3}{0.5} = 0.12\ \Omega$$

• Two identical cells, $\varepsilon = 5\ \text{V}$, $r = 2\ \Omega$, in parallel across $R = 10\ \Omega$: $\varepsilon_{eq} = 5\ \text{V}$, $r_{eq} = 1\ \Omega$, so $I = 5/11\ \text{A}$ and the voltage across $R$ is $50/11 \approx 4.55\ \text{V}$.
• Three cells of the same emf $E$ but internal resistances $2r$, $3r$ and $6r$ in parallel across $R$: $1/r_p = 1/2r + 1/3r + 1/6r = 1/r$, so $r_p = r$; the equivalent emf is $E$ because all three emfs are equal. Then $I = E/(R+r)$ and the terminal potential difference is $V = E - Ir = ER/(R+r)$.
• Same current in series and in parallel. For $n$ identical cells, $n\varepsilon/(R+nr) = \varepsilon/(R+r/n)$ gives $nR + r = R + nr$, i.e. $(n-1)R = (n-1)r$, so $R = r$.
• Three identical cells in series, from a $V$–$I$ plot running from $V = 6\ \text{V}$ at $I = 0$ to $V = 0$ at $I = 1\ \text{A}$: the intercept is the total emf $3\varepsilon = 6\ \text{V}$, so $\varepsilon = 2\ \text{V}$; the magnitude of the slope is the total internal resistance $3r = 6\ \Omega$, so $r = 2\ \Omega$.
• Two equal emfs in series with $r_2 > r_1$ across $R$: the terminal p.d. of the second source is zero when $\varepsilon = Ir_2$, i.e. $I = \varepsilon/r_2$. But $I = 2\varepsilon/(R+r_1+r_2)$, so $R + r_1 + r_2 = 2r_2$ and $R = r_2 - r_1$.
• A 10 V cell of negligible internal resistance connected in parallel across a 200 V battery of internal resistance $38\ \Omega$: the p.d. across each branch is 10 V, so $200 - 38I = 10$, giving $I = 5\ \text{A}$.

Standard results

• Series: $\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$, $r_{eq} = r_1 + r_2$; reversed cell subtracts its emf.
• Parallel: $\varepsilon_{eq} = (\varepsilon_1r_2 + \varepsilon_2r_1)/(r_1+r_2)$, $r_{eq} = r_1r_2/(r_1+r_2)$.
• General parallel: $1/r_{eq} = \Sigma(1/r_i)$ and $\varepsilon_{eq}/r_{eq} = \Sigma(\varepsilon_i/r_i)$.
• $n$ identical cells: $I_{\text{series}} = n\varepsilon/(R+nr)$, $I_{\text{parallel}} = \varepsilon/(R+r/n)$; equal when $R = r$.