Question

Define 'emf' of a cell. Derive expression for the equivalent emf and the equivalent internal resistance of two cells connected in series.

Hint:

Emf is the terminal voltage in open circuit. For cells in series add the terminal voltages, which gives E1 + E2 and r1 + r2.

Answer 1

Step 1: Define emf:
The emf of a cell is the potential difference between its two terminals when no current is drawn from it (open circuit).
It also equals the work done by the cell per unit charge in carrying the charge round the circuit.

Step 2: Set up the series combination:
Cell 1 has emf \( E_1 \) and internal resistance \( r_1 \). Cell 2 has emf \( E_2 \) and internal resistance \( r_2 \).
Join the negative terminal of cell 1 to the positive terminal of cell 2. Call the free positive terminal A, the junction B, and the free negative terminal C. A current I flows through both cells.

Step 3: Terminal voltages:
For cell 1: \( V_{AB} = V_A - V_B = E_1 - I r_1 \).
For cell 2: \( V_{BC} = V_B - V_C = E_2 - I r_2 \).

Step 4: Add them:
\[ V_{AC} = V_{AB} + V_{BC} = (E_1 + E_2) - I (r_1 + r_2) \]
Compare with a single cell: \( V = E_{eq} - I r_{eq} \).
\[ E_{eq} = E_1 + E_2, \qquad r_{eq} = r_1 + r_2 \]

Final Answer:
Equivalent emf = \( E_1 + E_2 \); equivalent internal resistance = \( r_1 + r_2 \). \[ \boxed{E_{eq} = E_1 + E_2,\ r_{eq} = r_1 + r_2} \]