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Q.

Two cells of emf’s E₁ and E₂ are connected in series in a circuit. Let r₁ and r₂ be the internal resistance of the cells. Then the current through the circuit is

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a

$\frac{2 (E_{1} + E_{2})}{(r_{1} + r_{2})}$

b

$\frac{E_{1} + E_{2}}{(r_{1} + r_{2})}$

c

$\frac{E_{1} - E_{2}}{(r_{1} + r_{2})}$

d

$\frac{E_{1} + E_{2}}{(r_{1} - r_{2})}$

answer is A.

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Detailed Solution

detailed_solution_thumbnail

By applying kirchhoff's voltage law

$E_{1} = {ir}_{1}; E_{2} = {ir}_{2}; E_{1} + E_{2} = i (r_{1} + r_{2}) \Rightarrow i = \frac{E_{1} + E_{2}}{r_{1} + r_{2}}$

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