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Simple Circuit

Two cells of ...

Question

Two cells of unequal emfs, $E_{1}$ and $E_{2}$ , the 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 with 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 grater than its emf

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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 one cell with continuously supply energy to the other
B the potential difference across both the cells will be equal
C the potential difference across one cell will be grater than its emf
Let $i$ be the current flow in clockwise direction in the circuit .
thus, $E_{2} - E_{1} = i (r_{1} + r_{2})$
or $i = \frac{E_{2} - E_{1}}{r_{1} + r_{2}}$
now, $V_{A} - V_{B} = E_{2} + i r_{2} = E_{2} + \frac{(E_{2} - E_{1}) r_{2}}{r_{1} + r_{2}} = \frac{E_{2} r_{1} + E_{2} r_{2} + E_{2} r_{2} - E_{1} r_{2}}{r_{1} + r_{2}} = \frac{E_{2} r_{1} + 2 E_{2} r_{2} - E_{1} r_{2}}{r_{1} + r_{2}}$
This is also potential across each cell.
Thus, $V_{A} - V_{B} > E_{2}$
Current flows in the cell of emf $E_{2}$ from the positive plate to the negative plate inside the cell and hence it absorbs energy.
ans :A,B,C

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

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