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What is the difference between ‘emf’ and ‘terminal voltage’ of a cell?

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In parallel combinations, use current-based equations and equate terminal voltages. The weighted average formula for EMF ensures voltage consistency across branches.
Updated On: Jan 13, 2026
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Solution and Explanation

(a) Electromotive Force and Terminal Voltage

Electromotive Force (EMF):
EMF is defined as the maximum potential difference across a cell's terminals when no current is drawn. It signifies the total energy per unit charge supplied by the cell.

Terminal Voltage:
Terminal voltage is the potential difference across a cell's terminals when it is supplying current. Due to internal resistance \( r \), a voltage drop occurs within the cell. Therefore:

\[ \text{Terminal voltage} = \text{EMF} - Ir \]

Relationship:

\[ \text{EMF} \geq \text{Terminal voltage} \quad (\text{Equality holds when } I = 0) \]

(b) Derivation for Two Cells in Parallel

Given: Two cells with EMFs \( E_1 \) and \( E_2 \), and internal resistances \( r_1 \) and \( r_2 \) respectively, connected in parallel.

Objective: To derive the expressions for the equivalent EMF \( E \) and equivalent internal resistance \( r \).

Derivation:
For parallel connections, the terminal voltages of the cells are equal. Let:

\[ E_1 - I_1 r_1 = E_2 - I_2 r_2 = V \]

Let the total current be \( I = I_1 + I_2 \). For the equivalent cell, the relationship is:

\[ V = E - Ir \]

From the current equations:

\[ I_1 = \frac{E_1 - V}{r_1}, \quad I_2 = \frac{E_2 - V}{r_2} \]

The total current is the sum of individual currents:

\[ I = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2} \Rightarrow I = \frac{E_1}{r_1} + \frac{E_2}{r_2} - V\left(\frac{1}{r_1} + \frac{1}{r_2} \right) \]

Substitute this expression for \( I \) into the equivalent cell equation \( V = E - Ir \):

\[ V = E - r\left( \frac{E_1}{r_1} + \frac{E_2}{r_2} - V\left( \frac{1}{r_1} + \frac{1}{r_2} \right) \right) \]

Solving for \( E \) and \( r \) yields:

Equivalent EMF:

\[ E = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} \]

Equivalent Internal Resistance:

\[ \frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} \]

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