Ncert Exercises & Solutions

The circuit in the figure shows two cells connected in opposition to each other. Cell $E_{1}$ $E_{1}$ is of emf 6V and internal resistance 2 $Ω$ $Ω$ ; the cell $E_{2}$ $E_{2}$ is of emf 4V and internal resistance 8 $Ω$ $Ω$ . Find the potential difference between points A and B.

Hint: The potential difference between A and B depends on the current in the circuit.

Step 1: Find the current in the circuit.

Effective resistance =

2 Ω + 8 Ω = 10 Ω

$2 Ω + 8 Ω = 10 Ω$ and effective emf of two cells= 6-4 = 2V

So the electric current is given by;

I = \frac{6 - 4}{2 + 8} = 0 .2 A

$I = \frac{6 - 4}{2 + 8} = 0 .2 A$ along anti-clockwise direction, since

E_{1}

$E_{1}$ > E.

Step 2: Find the potential difference between A and B.
The direction of the flow of current is always from high potential to low potential. Therefore;

$∴ V_{B} > V_{A}$

$\Rightarrow V_{B} - 4 V - (0 .2) \times 8 = V_{A}$

$∴ V_{B} - V_{A} = 3 .6 V$

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