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A battery B-1 is made by using two cells of emf $E_{1}$ and $E_{2}$ internal resistance $r_{1} and r_{2}$ respectively. Another battery B-2 is made by using different cells of emf $E_{3}$ and $E_{4}$ and internal resistance $r_{3} and r_{4}$ respectively. Now they are connected with points P and Q as shown in the figure. All cells are rechargeable. Now switch is closed at t=0, then choose the correct option(s).

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Rate of loss in chemical energy of one battery is more than the rate of gain in chemical energy of the other

Rate of heat dissipation in both batteries is equal.

The potential difference across both the battery will be equal

One battery will charge the other battery.

answer is A, C, D.

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Detailed Solution

$i = \frac{| E_{1} + E_{2} - E_{3} - E_{4} |}{r_{1} + r_{2} + r_{3} + r_{4}}$

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