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Two cells of emf E₁ and E₂ are connected to two resistors R₁ and R₂ as shown. If E₂ is short circuited then currents through R₁ and R₂ are

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a

$\frac{E_{1}}{R_{1}}, 0$

b

$0, \frac{E_{1}}{R_{2}}$

c

$0, \frac{E_{1}}{R_{1}}$

d

$\frac{E_{1}}{R_{2}}, 0$

answer is C.

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Detailed Solution

After short circuiting, R₂ becomes meaningless. Current will not flow through R₂.

So, current will be $\frac{E_{1}}{R_{1}}$

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