A cell E1 of emf 6V and internal resistance 2Ω is connected with another cell E2 of emf 4 V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is :

Current I=6−410=15A As the 4 v cell is in charging, P.D across x and y is 4 + (8x1/5) =5.6 V

A cell E1 of emf 6V and internal resistance 2Ω is connected with another cell E2 of emf 4 V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is :

Current I=6−410=15A As the 4 v cell is in charging, P.D across x and y is 4 + (8x1/5) =5.6 V

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A cell E₁of emf 6V and internal resistance 2 $Ω$ is connected with another cell E₂ of emf 4 V and internal resistance 8 $Ω$ (as shown in the figure). The potential difference across points X and Y is :

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10.0 V

3.6 V

5.6 V

2.0 V

answer is C.

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Detailed Solution

$Current I = \frac{6 - 4}{10} = \frac{1}{5} A$

As the 4 v cell is in charging, P.D across x and y is 4 + (8x1/5) =5.6 V

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A cell E1 of emf 6V and internal resistance 2Ω is connected with another cell E2 of emf 4 V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is :

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A cell E₁of emf 6V and internal resistance 2 $Ω$ is connected with another cell E₂ of emf 4 V and internal resistance 8 $Ω$ (as shown in the figure). The potential difference across points X and Y is :