Under what condition current passing through the resistance $R$ can be increased by short circuiting the battery of emf $E_{2}$ . The internal resistances of the two batteries are $r_{1}$ and $r_{2}$ respectively.

E_{2} r_{1} > E_{1} (R + r_{2})

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E_{1} r_{2} > E_{2} (R + r_{1})

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E_{2} r_{2} > E_{1} (R + r_{2})

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E_{1} r_{1} > E_{2} (R + r_{1})

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Solution

The correct option is B $E_{1} r_{2} > E_{2} (R + r_{1})$
Comparing current through both batteries,

$I = \frac{E_{1}}{R + r_{1}} = \frac{E_{2}}{R + r_{1} + r_{2}}$

Hence,

$\frac{E_{1}}{R + r_{1}} = \frac{E_{2}}{R + r_{1} + r_{2}}$

$E_{1} (R + r_{1} + r_{2}) = E_{2} (R + r_{1})$

$E_{1} R + E_{1} r_{1} + E_{1} r_{2} = E_{2} R + E_{2} r_{1}$

$E_{1} (R + r_{1}) + E_{1} r_{2} = E_{2} R + E_{2} r_{1}$

$E_{1} r_{2} = E_{2} (R + r_{1}) - E_{1} (R + r_{1})$

Therefore,

$E_{1} r_{2} > E_{2} (R + r_{1})$

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Under what condition current passing through the resistance R can be increased by short circuiting the battery of emf E2. The internal resistances of the two batteries are r1 and r2 respectively.

The correct option is B E1r2>E2(R+r1)Comparing current through both batteries,I=E1R+r1=E2R+r1+r2Hence,E1R+r1=E2R+r1+r2E1(R+r1+r2)=E2(R+r1)E1R+E1r1+E1r2=E2R+E2r1E1(R+r1)+E1r2=E2R+E2r1E1r2=E2(R+r1)−E1(R+r1)Therefore,E1r2>E2(R+r1)

Under what condition current passing through the resistance $R$ can be increased by short circuiting the battery of emf $E_{2}$ . The internal resistances of the two batteries are $r_{1}$ and $r_{2}$ respectively.