Question

Find the equivalent emf and the equivalent resistance between points A and B for two cells arranged as shown in the figure.

Hint:

The cells are in parallel with opposite polarity, so the loop emf adds up to 9 V. Use \(I = 9/1.8\) to find \(V_{AB}\), and the parallel rule for the resistance.

Answer 1

Step 1: Read the circuit:
Two cells are joined between points A and B. The cell of 6 V (internal resistance 1 ohm) has its positive plate at A. The cell of 3 V (internal resistance 0.8 ohm) has its positive plate at B. So they are in parallel, but with opposite polarity.

Step 2: Find the current in the loop:
With A and B open (nothing else connected), a current \(I\) flows round the loop through both cells. Both emfs push the current in the same sense around the loop, so they add.
\[ I = \frac{6 + 3}{1 + 0.8} = \frac{9}{1.8} = 5\ \text{A} \]

Step 3: Find the emf between A and B:
The equivalent emf is the potential difference \(V_{AB}\) on open circuit. Using the 6 V cell:
\[ V_{AB} = 6 - I \times 1 = 6 - 5 = 1\ \text{V} \]
Check with the 3 V cell: \(V_{AB} = -3 + I \times 0.8 = -3 + 4 = 1\ \text{V}\). Both give the same value.

Step 4: Find the equivalent resistance:
The internal resistances are in parallel between A and B:
\[ r_{eq} = \frac{1 \times 0.8}{1 + 0.8} = \frac{0.8}{1.8} = \frac{4}{9}\ \Omega \approx 0.44\ \Omega \]

Final Answer:
The pair acts as a single cell of emf 1 V with A as the positive terminal, and internal resistance 0.44 ohm. \[ \boxed{E_{eq} = 1\ \text{V},\quad r_{eq} = \frac{4}{9}\ \Omega \approx 0.44\ \Omega} \]