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Home » The circuit in the figure shows two cells connected in opposition to each other. Cell E1 is of emf 6V and internal resistance 2Ω; the cell E2 is of emf 4V and internal resistance 8 Ω. Find the potential difference between the points A and B.
The circuit in the figure shows two cells connected in opposition to each other. Cell E1 is of emf 6V and internal resistance 2Ω; the cell E2 is of emf 4V and internal resistance 8 Ω. Find the potential difference between the points A and B.
CBSE | Current Electricity | Lakhmir Singh | Physics
The circuit in the figure shows two cells connected in opposition to each other. Cell E1 is of emf 6V and internal resistance 2Ω; the cell E2 is of emf 4V and internal resistance 8 Ω. Find the potential difference between the points A and B.

Applying Ohm’s law, equivalent emf of the two cells = 6 – 4 = 2V

Equivalent resistance = 2 + 8 = 10 Ω

Electric current, I = 6-4/2+8 = 0.2A

When the loop is considered in the anti-clockwise direction, E1 > E2 which means VB > VA

Therefore, VB – VA = 3.6 V

i

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