Question:medium

Two cells of EMFs \( E_1 \) and \( E_2 \), and internal resistances \( r_1 \) and \( r_2 \), are connected in parallel. Derive an expression for the EMF and internal resistance of the equivalent cell.

Updated On: Jan 13, 2026
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Solution and Explanation

Let \( E \) be the equivalent electromotive force (EMF) and \( r \) be the equivalent internal resistance of the two cells connected in parallel. The terminal voltages across both cells are equal. This can be expressed as:

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

Here, \( V \) denotes the terminal voltage, \( I_1 \) and \( I_2 \) are the currents through each cell, and \( r_1 \) and \( r_2 \) are their respective internal resistances.

The total current \( I \) from the equivalent cell is the sum of the individual currents:

\[ I = I_1 + I_2 \]

Expressing \( I_1 \) and \( I_2 \) in terms of \( V \):

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

Substituting these into the total current equation:

\[ I = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2} \]

Rearranging to express \( I \) as a function of \( V \):

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

Applying Ohm's law to the equivalent cell, \( V = E - I r \). Substituting the expression for \( I \):

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

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

1. Equivalent EMF:

The equivalent EMF is calculated as:

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

2. Equivalent Internal Resistance:

The reciprocal of the equivalent internal resistance is:

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

Therefore, the equivalent internal resistance is \( r = \frac{r_1 r_2}{r_1 + r_2} \).

Summary of derived values:

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