Internal Resistance of a Cell 

1.0What is Internal Resistance?

Internal resistance (r) is the opposition to the flow of current within a cell or battery itself. It's an inherent property that arises from the materials of the electrodes and the electrolyte inside the cell. When a current flows, some of the energy supplied by the cell is dissipated as heat due to this internal resistance, causing a drop in the voltage delivered to the external circuit.

The presence of internal resistance means that a real cell is not a perfect voltage source. It can be modeled as an ideal battery with an electromotive force (EMF) (E) in series with a small resistor (r).

2.0Internal Resistance of a Cell Formula

The internal resistance of a cell can be calculated using the relationship between its EMF (EE), terminal voltage (VV), and the current (II) flowing through the external circuit:

Resistance offered by the electrolyte of the cell when an electric current flows through it is known as internal resistance.

Dependency of internal resistance r :

• Distance between two electrodes increases  r increases

• Area dipped in electrolyte increases  r decreases

• Concentration of electrolyte increases  r increases

• Temperature increases  r decreases

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

Where:

• r = internal resistance of the cell (Ω)
• E = electromotive force (EMF) of the cell (V)
• V = terminal voltage across the cell (V)
• I = current flowing through the external circuit (A)

3.0Internal Resistance of a Cell Derivation

The internal resistance of a cell (r) is the opposition offered by the cell's electrodes and electrolyte to the flow of charge. A real cell can be modeled as an ideal source of electromotive force (EMF), E, in series with an internal resistance, r. This model helps explain why the voltage across a battery's terminals, known as terminal voltage (V), drops when it's connected to an external circuit.

Terminal potential difference (V)

• When current is drawn through a cell or current is supplied to it then the potential difference across its terminals is called terminal voltage.
• When I current is drawn from cell then terminal voltage V is less than its e.m.f i.e., V=E−Ir

When no current is drawn from the cell (an open circuit), the terminal voltage is equal to the cell's EMF, as no voltage is dropped across the internal resistance.

$V=\mathcal{E}\text{ (for }I=0\text{)}$

When the cell is connected to an external circuit with a total external resistance R, a current I flows. According to Ohm's law, the potential difference across the external resistance is the terminal voltage.

V=IR

The total resistance in the circuit is the sum of the external resistance and the internal resistance: 

$R_{total}=R+r$

Using Ohm's law for the entire circuit, the current flowing is:

$I=\frac{\text{Total EMF}}{\text{Total Resistance}}=\frac{\mathcal{E}}{R+r}$

To derive the relationship for terminal voltage, we rearrange this equation:

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

$IR+Ir=\mathcal{E}$

Since V=IR, we can substitute it into the equation:

$V+Ir=\mathcal{E}$

This gives the final equation for the terminal voltage:

$V=\mathcal{E}-Ir$

This is the derivation of the terminal voltage relationship. From this, we can also derive the formula for internal resistance:

$Ir=\mathcal{E}-V$

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

This shows that the internal resistance is the ratio of the "lost volts" (E−V) to the current flowing through the circuit.

(b) When cell is getting charged :

Current inside the cell is from anode to cathode.

$\text{Current }I=\frac{V-E}{r}⟹V=E+Ir$

During charging terminal potential difference is greater than the emf of the cell.

(c) When cell is in the open circuit:
In open circuit ( $R=\infty$ )

$I=\frac{E}{R+r}=0\Rightarrow V=E$

In open circuit, terminal potential difference is equal to emf and is the maximum potential difference which a cell can provide.

(d) When cell is short circuited:
In short circuit ( R = 0 )

$I=\frac{E}{R+r}=\frac{E}{r}\text{and}V=IR=0$

Note: In short circuit, current from the cell is maximum and terminal potential difference is zero.

4.0Factors Affecting Internal Resistance

The value of a cell's internal resistance depends on several factors:

• Nature of the Electrodes: Different electrode materials have different resistance to current flow.
• Nature of the Electrolyte: The resistivity of the electrolyte solution contributes to the internal resistance.
• Concentration of the Electrolyte: A higher concentration can lead to lower internal resistance, but only up to a point.
• Temperature: Generally, internal resistance decreases with an increase in temperature because the mobility of ions in the electrolyte increases.
• Age of the Cell: As a cell ages, the electrolyte and electrodes can degrade, leading to an increase in internal resistance.

5.0Terminal Voltage and EMF

Terminal Voltage and EMF

Terminal Potential Difference (TPD or V)
The potential difference between the two electrodes of a cell in a closed circuit i.e., when current is being drawn from the cell is called terminal potential difference.

Electromotive force : (E.M.F)
Definition I : Electromotive force is the capability of the system to make the charge flow.
Definition II : It is the work done by the battery for the flow of 1 coulomb charge from lower potential terminal to higher potential terminal inside the battery.
Definition III : Electromotive force of a cell is equal to potential difference between its terminals when no current is passing through the circuit.

• Electromotive Force (EMF) (E): The EMF is the total energy per unit charge provided by the cell. It's the voltage across the cell terminals when no current is being drawn from it (i.e., in an open circuit). It's the maximum potential difference the cell can provide.
• Terminal Voltage (V): The terminal voltage is the actual voltage measured across the terminals of the cell when it is connected to a circuit and a current is flowing. Due to the voltage drop across the internal resistance, the terminal voltage is always less than the EMF when the cell is discharging.

At the time of charging a cell, when current is being supplied to it, the terminal voltage is greater than the emf E.

$V=E+Ir\text{So}V>E$

Series combination is useful when internal resistance of the cell is less than external resistance.
Parallel combination is useful when internal resistance of the cell is greater than external resistance.
Internal resistance of an ideal cell = 0
If external resistance is zero, then current delivered by the battery is maximum.

6.0Relation Between Internal Resistance, EMF, and Terminal Voltage

When a cell with EMF E and internal resistance r is connected to an external resistance R, the current (I) flowing in the circuit is:

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

The terminal voltage (V) across the external resistance is given by Ohm's Law:

V=IR

Substituting the expression for I, we get:

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

This can be rearranged to find the voltage drop across the internal resistance, often called the "lost volts."

$V=\mathcal{E}-Ir$

This is the most common and useful formula. It shows that the terminal voltage (V) is equal to the EMF (E) minus the voltage drop (Ir) across the internal resistance.

From this, the internal resistance can be expressed as:

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

7.0Solved JEE-Level Problems

Problem 1: A cell of EMF 2 V and internal resistance 0.1Ω is connected to a resistor of 3.9Ω. Find the current drawn from the cell and its terminal voltage.

Solution:

$\mathcal{E}=2\text{ V}$

$r=0.1\Omega$

$R=3.9\Omega$

Current (I):

$I=\frac{\mathcal{E}}{R+r}=\frac{2}{3.9+0.1}=\frac{2}{4.0}=0.5\text{ A}$

Terminal Voltage (V):

$V=IR=(0.5)(3.9)=1.95\text{ V}$

$V=\mathcal{E}-Ir:$

$V=2-(0.5)(0.1)=2-0.05=1.95\text{ V}$

Problem 2: A cell's terminal voltage is 1.8 V when it is supplying a current of 0.5 A. Its EMF is 2.2 V. What is the internal resistance of the cell?

Solution:

$V=1.8\text{ V}$

$I=0.5\text{ A}$

$\mathcal{E}=2.2\text{ V}$

$V=\mathcal{E}-Ir:$

1.8 = 2.2 - (0.5)r

0.5r = 2.2 - 1.8 = 0.4

$r=\frac{0.4}{0.5}=0.8\Omega$

Problem 3: A cell of e.m.f. E is connected with an external resistance R, then p.d. across cell is V. The internal resistance of cell will be

$(A)\frac{(E-V)R}{E}(B)\frac{(E-V)R}{V}(C)\frac{(V-E)R}{V}(D)\frac{(V-E)R}{E}$

Solution:

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

Potential difference across RR is same as that of battery.

V = IR

$V=\frac{ER}{(R+r)}⟹r=\frac{ER}{V}-R$