Question

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.

Answer 1

Let the equivalent EMF of the two cells be \( E \), and the equivalent internal resistance be \( r \). Since the cells are connected in parallel, the terminal voltages across both cells are equal. We can write:

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

Where \( V \) is the terminal voltage across both cells, \( I_1 \) and \( I_2 \) are the currents through the two cells, and \( r_1 \) and \( r_2 \) are their respective internal resistances.

The total current \( I \) supplied by the equivalent cell is the sum of the currents from the two cells:

\[ I = I_1 + I_2 \]

Now, expressing the currents \( I_1 \) and \( I_2 \) in terms of the terminal voltage \( V \), we get:

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

Thus, the total current is:

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

Now, rearrange the equation to get the total current \( I \) in terms of the terminal voltage \( V \):

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

Now, using Ohm’s law for the equivalent cell, we know that:

\[ V = E - I r \]

Substituting the expression for \( I \) into this equation:

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

Now solve for \( E \) and \( r \), we get:

1. Equivalent EMF:

The equivalent EMF \( E \) is given by:

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

2. Equivalent Internal Resistance:

The equivalent internal resistance \( r \) is given by:

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

Thus, the equivalent internal resistance of the two cells connected in parallel is \( r = \frac{r_1 r_2}{r_1 + r_2} \).

Summary of the results: