Question

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

Hint:

In parallel combinations, use current-based equations and equate terminal voltages. The weighted average formula for EMF ensures voltage consistency across branches.

Answer 1

(a) EMF and Terminal Voltage 

EMF (Electromotive Force):
EMF is the maximum potential difference between the terminals of a cell when no current is being drawn from it. It represents the total energy supplied per unit charge by the cell.

Terminal Voltage:
Terminal voltage is the potential difference between the terminals of the cell when it is supplying current. Due to internal resistance \( r \), some voltage is lost inside the cell. Hence:

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

Difference:

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

(b) Derivation for Two Cells in Parallel

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

Objective: Derive the expression for equivalent EMF \( E \) and equivalent internal resistance \( r \).

Solution:
Since the cells are connected in parallel, their terminal voltages must be equal. Let:

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

Let the total current be \( I = I_1 + I_2 \), and for the equivalent cell:

\[ V = E - Ir \]

From the current expressions:

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

Total current becomes:

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

Substitute into \( V = E - Ir \):

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

Solve for \( E \) and \( r \), and we get:

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:

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