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Mixed Combination of Cells

Two cells of ...

Question

Two cells of unequal e.m.fs. $E_{1}$ and $E_{2}$ and internal resistance $r_{1}$ and $r_{2}$ are joined as shown. $V_{A}$ and $V_{B}$ are the potentials at A and B respectively:

one cell will continuously supply energy to the other

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the potential difference across both the cells will be equal

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the potential difference across one cell will be greater than its e.m.f.

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V_{A} - V_{B} = \frac{(E_{1} R_{2} + E_{2} R_{1})}{(r_{1} + r_{2})}

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Solution

The correct options are
A the potential difference across both the cells will be equal
B one cell will continuously supply energy to the other
C the potential difference across one cell will be greater than its e.m.f.
Let $E_{1} < E_{2}$

Hence current in the circuit becomes $I = \frac{E_{1} - E_{2}}{r_{1} + r_{2}}$

Potential difference across each cell $V_{A} - V_{B} = E_{2} + I r_{2}$

$⟹ V_{A} - V_{B} = \frac{E_{2} r_{1} + E_{1} r_{2}}{r_{1} + r_{2}}$
Current flows in $E_{2}$ from positive plate to the negative plate inside the

cell and hence it absorbs the energy.

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Two cells of unequal e.m.fs. E1 and E2 and internal resistance r1 and r2 are joined as shown. VA and VB are the potentials at A and B respectively:

Two cells of unequal e.m.fs. $E_{1}$ and $E_{2}$ and internal resistance $r_{1}$ and $r_{2}$ are joined as shown. $V_{A}$ and $V_{B}$ are the potentials at A and B respectively: