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Cells, EMF, and Internal Resistance



Cells, Emf ($ \mathcal{E} $), Internal Resistance ($ r $)

An electric cell or a battery (a combination of cells) is a source of electrical energy. Its primary function is to maintain a continuous flow of charge, i.e., an electric current, in a closed circuit by providing a potential difference across its terminals. This potential difference drives the charge carriers through the circuit components. Cells convert chemical energy into electrical energy.


Electromotive Force (EMF), $\mathcal{E}$

The electromotive force (EMF) of a cell is the maximum potential difference that the cell can provide. More precisely, it is defined as the energy supplied by the cell per unit charge when no current is flowing through the cell itself (i.e., in an open circuit). It is a measure of the energy that the source can impart to each unit of charge. EMF is characteristic of the cell's chemistry.

The symbol for EMF is usually $\mathcal{E}$ (epsilon) or sometimes E. The unit of EMF is the same as potential difference, which is the volt (V).

Think of EMF as the "voltage generated" by the chemical reactions inside the cell, or the total "electrical pressure" available from the source before any current is drawn from it through an external circuit.


Internal Resistance ($r$)

Real electric cells and batteries are not perfect. The materials within the cell itself (the electrolyte, electrodes, etc.) offer some opposition to the flow of charge through them. This opposition within the cell is called its internal resistance ($r$).

Internal resistance depends on the nature of the electrolyte and electrodes, their area, the distance between the electrodes, and the temperature. For example, the internal resistance of a fresh dry cell is usually low, but it increases as the cell ages or if the temperature drops.

The unit of internal resistance is the ohm ($\Omega$).


Terminal Voltage ($V$)

When a cell is connected in a closed circuit and current ($I$) is drawn from it, the potential difference across the terminals of the cell is called the terminal voltage ($V$).

Due to the internal resistance ($r$), some potential drop occurs within the cell itself when current flows. This internal potential drop is given by $Ir$ (according to Ohm's Law applied to the internal resistance).

The terminal voltage ($V$) available to the external circuit is the EMF ($\mathcal{E}$) minus this internal potential drop ($Ir$).

$ V = \mathcal{E} - I r $

This equation shows that the terminal voltage ($V$) is always less than the EMF ($\mathcal{E}$) when current ($I$) is flowing out of the positive terminal of the cell (discharging). When no current is flowing ($I=0$, open circuit), the terminal voltage equals the EMF ($V = \mathcal{E}$).

If the cell is being charged, current flows into the positive terminal. In this case, the internal potential drop $Ir$ is added to the EMF, and the terminal voltage across the cell is $V = \mathcal{E} + Ir$.

The terminal voltage is the potential difference measured across the external circuit components connected to the cell. If an external resistor of resistance $R$ is connected across the terminals of the cell, the terminal voltage $V$ is also given by Ohm's Law applied to the external circuit:

$ V = I R $

Equating the two expressions for $V$:

$ IR = \mathcal{E} - I r $

Rearranging to find the current $I$ flowing in the circuit:

$ IR + I r = \mathcal{E} $

$ I (R + r) = \mathcal{E} $

$ I = \frac{\mathcal{E}}{R + r} $

This formula gives the total current flowing in a circuit containing a single cell with EMF $\mathcal{E}$ and internal resistance $r$ connected to an external resistance $R$. The total resistance in the circuit is the sum of the external resistance and the internal resistance ($R+r$).


Graphical Representation (V vs I)

The relationship between terminal voltage ($V$) and current ($I$) for a cell is $V = \mathcal{E} - Ir$. This is the equation of a straight line if $V$ is plotted on the y-axis and $I$ on the x-axis.

Graph of Terminal Voltage (V) vs Current (I) for a real cell

Graph of Terminal Voltage (V) vs Current (I) for a cell.


Example 1. A battery has an EMF of 12 V and an internal resistance of 2 $\Omega$. If it is connected to an external resistor of 10 $\Omega$, calculate (a) the current flowing through the circuit and (b) the terminal voltage across the battery.

Answer:

Given:

EMF of the battery, $\mathcal{E} = 12 \, V$

Internal resistance, $r = 2 \, \Omega$

External resistance, $R = 10 \, \Omega$

(a) To find the current ($I$) flowing through the circuit, we use the formula $I = \frac{\mathcal{E}}{R + r}$.

Substitute the given values:

$ I = \frac{12 \, V}{10 \, \Omega + 2 \, \Omega} = \frac{12 \, V}{12 \, \Omega} = 1 \, A $

The current flowing through the circuit is 1 Ampere.

(b) To find the terminal voltage ($V$) across the battery, we can use either $V = \mathcal{E} - Ir$ or $V = IR$. Using $V = \mathcal{E} - Ir$:

$ V = 12 \, V - (1 \, A) \times (2 \, \Omega) $

$ V = 12 \, V - 2 \, V = 10 \, V $

Using $V = IR$ (potential drop across the external resistor):

$ V = (1 \, A) \times (10 \, \Omega) = 10 \, V $

Both methods give the same terminal voltage. The terminal voltage across the battery is 10 V, which is less than the EMF (12 V) due to the voltage drop across the internal resistance.



Cells In Series And In Parallel

Cells can be connected in series or parallel combinations to achieve desired voltage levels or current capabilities.


Cells In Series

When cells are connected in series, the positive terminal of one cell is connected to the negative terminal of the next cell, and so on. This arrangement increases the total EMF and the total internal resistance of the combination.

Diagram showing two cells connected in series to an external resistor

Two cells with EMFs $\mathcal{E}_1, \mathcal{E}_2$ and internal resistances $r_1, r_2$ connected in series.

Consider $n$ cells connected in series. Let the EMFs be $\mathcal{E}_1, \mathcal{E}_2, ..., \mathcal{E}_n$ and their internal resistances be $r_1, r_2, ..., r_n$. If connected such that they all aid each other (positive to negative, positive to negative, ...), the total or equivalent EMF ($\mathcal{E}_{eq}$) is the sum of the individual EMFs:

$ \mathcal{E}_{eq} = \mathcal{E}_1 + \mathcal{E}_2 + ... + \mathcal{E}_n $

$ \mathcal{E}_{eq} = \sum_{i=1}^{n} \mathcal{E}_i $

The total or equivalent internal resistance ($r_{eq}$) is the sum of the individual internal resistances:

$ r_{eq} = r_1 + r_2 + ... + r_n $

$ r_{eq} = \sum_{i=1}^{n} r_i $

If the combination is connected to an external resistance $R$, the current flowing through the circuit is:

$ I = \frac{\mathcal{E}_{eq}}{R + r_{eq}} = \frac{\sum \mathcal{E}_i}{R + \sum r_i} $


Identical Cells in Series

If $n$ identical cells, each with EMF $\mathcal{E}$ and internal resistance $r$, are connected in series:

$ \mathcal{E}_{eq} = n\mathcal{E} $

$ r_{eq} = nr $

The current in the external circuit with resistance $R$ is:

$ I = \frac{n\mathcal{E}}{R + nr} $

Advantage: The main advantage of connecting cells in series is to obtain a higher voltage. This is useful for applications requiring higher voltage.


Cells in Series with Reverse Polarity

If one or more cells in a series connection are connected with reverse polarity (positive terminal connected to positive terminal, or negative to negative), their EMFs will subtract from the total EMF. For example, if one cell with EMF $\mathcal{E}_j$ is reversed in a series of $n$ cells, the total EMF will be $\sum_{i=1}^{n} \mathcal{E}_i - 2\mathcal{E}_j$ (assuming the sum was initially of EMFs connected in the same direction). The internal resistances still add up: $r_{eq} = \sum r_i$.


Cells In Parallel

When cells are connected in parallel, all the positive terminals are connected to one point, and all the negative terminals are connected to another point. This arrangement maintains the voltage across the external circuit while potentially increasing the total current capacity and the overall life of the battery.

Diagram showing two cells connected in parallel to an external resistor

Two cells with EMFs $\mathcal{E}_1, \mathcal{E}_2$ and internal resistances $r_1, r_2$ connected in parallel.

Consider $n$ cells connected in parallel. Let the EMFs be $\mathcal{E}_1, \mathcal{E}_2, ..., \mathcal{E}_n$ and their internal resistances be $r_1, r_2, ..., r_n$.

Key characteristics of parallel connection:


Equivalent EMF and Internal Resistance in Parallel (General Case)

For a parallel combination of cells with EMFs $\mathcal{E}_1, \mathcal{E}_2, ..., \mathcal{E}_n$ and internal resistances $r_1, r_2, ..., r_n$, connected to an external resistance $R$, let the terminal voltage be $V$.

For each cell, the terminal voltage is $V = \mathcal{E}_i - I_i r_i$, so the current from the $i$-th cell is $I_i = \frac{\mathcal{E}_i - V}{r_i}$.

The total current $I = \sum I_i = \sum \frac{\mathcal{E}_i - V}{r_i} = \sum \left(\frac{\mathcal{E}_i}{r_i} - \frac{V}{r_i}\right)$.

$ I = \sum \frac{\mathcal{E}_i}{r_i} - V \sum \frac{1}{r_i} $

Rearranging this equation to the form $V = \mathcal{E}_{eq} - I r_{eq}$:

$ V \sum \frac{1}{r_i} = \sum \frac{\mathcal{E}_i}{r_i} - I $

$ V = \frac{\sum (\mathcal{E}_i/r_i)}{\sum (1/r_i)} - \frac{I}{\sum (1/r_i)} $

Comparing this with $V = \mathcal{E}_{eq} - I r_{eq}$, we get:

Equivalent EMF: $ \mathcal{E}_{eq} = \frac{\sum_{i=1}^{n} (\mathcal{E}_i/r_i)}{\sum_{i=1}^{n} (1/r_i)} $

Equivalent internal resistance: $ \frac{1}{r_{eq}} = \sum_{i=1}^{n} \frac{1}{r_i} $


Identical Cells in Parallel

If $n$ identical cells, each with EMF $\mathcal{E}$ and internal resistance $r$, are connected in parallel:

The equivalent internal resistance is given by:

$ \frac{1}{r_{eq}} = \frac{1}{r} + \frac{1}{r} + ... + \frac{1}{r} \quad (n \text{ terms}) $

$ \frac{1}{r_{eq}} = \frac{n}{r} \implies r_{eq} = \frac{r}{n} $

The equivalent EMF:

$ \mathcal{E}_{eq} = \frac{\sum (\mathcal{E}/r)}{\sum (1/r)} = \frac{n(\mathcal{E}/r)}{n(1/r)} = \mathcal{E} $

So, for identical cells in parallel, the equivalent EMF is equal to the EMF of a single cell. The internal resistance decreases by a factor of $n$.

The current in the external circuit with resistance $R$ is:

$ I = \frac{\mathcal{E}_{eq}}{R + r_{eq}} = \frac{\mathcal{E}}{R + r/n} = \frac{n\mathcal{E}}{nR + r} $

Advantages:

Note: Connecting cells with different EMFs in parallel can lead to circulating currents within the parallel loop itself, potentially wasting energy and damaging the cells, especially if their internal resistances are low. It is generally recommended to connect cells of the same type and charge level in parallel.


Example 2. Four identical cells, each of EMF 1.5 V and internal resistance 0.5 $\Omega$, are connected in (a) series and (b) parallel to an external resistor of 5 $\Omega$. Calculate the current flowing in the external circuit in each case.

Answer:

Given: Number of cells $n = 4$. Each cell has EMF $\mathcal{E} = 1.5 \, V$ and internal resistance $r = 0.5 \, \Omega$. External resistance $R = 5 \, \Omega$.

(a) Cells in Series:

Equivalent EMF, $\mathcal{E}_{eq} = n\mathcal{E} = 4 \times 1.5 \, V = 6.0 \, V$

Equivalent internal resistance, $r_{eq} = nr = 4 \times 0.5 \, \Omega = 2.0 \, \Omega$

Current in the external circuit, $I_{series} = \frac{\mathcal{E}_{eq}}{R + r_{eq}}$

$ I_{series} = \frac{6.0 \, V}{5 \, \Omega + 2.0 \, \Omega} = \frac{6.0 \, V}{7.0 \, \Omega} $

$ I_{series} \approx 0.857 \, A $

The current flowing in the external circuit when cells are in series is approximately 0.857 A.

(b) Cells in Parallel:

Equivalent EMF, $\mathcal{E}_{eq} = \mathcal{E} = 1.5 \, V$

Equivalent internal resistance, $r_{eq} = r/n = 0.5 \, \Omega / 4 = 0.125 \, \Omega$

Current in the external circuit, $I_{parallel} = \frac{\mathcal{E}_{eq}}{R + r_{eq}}$

$ I_{parallel} = \frac{1.5 \, V}{5 \, \Omega + 0.125 \, \Omega} = \frac{1.5 \, V}{5.125 \, \Omega} $

$ I_{parallel} \approx 0.293 \, A $

The current flowing in the external circuit when cells are in parallel is approximately 0.293 A.

Observation: In this example, connecting identical cells in series gives a significantly higher current because the external resistance (5 $\Omega$) is much larger than the internal resistance of a single cell (0.5 $\Omega$). If the external resistance were very small compared to the internal resistance, parallel connection would likely yield a higher current (closer to $n\mathcal{E}/r$).