Under what conditions 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.

Question

Under what conditions 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. (A) E2 r1>E1(R+r2) (B) E1 r2< E2(r1+R) (C) E2 r2< E(R+r2) (D) E1 r1>E2(r1+R)

Tardigrade Admin · Accepted Answer

The current through the circuit before the battery of emf

E_{2}

is short circuited, is

i_{1} = \frac{E _{1} + E _{2}}{r _{1} + r _{2} + R}

...(i)
After short circuiting the battery of emf

E_{2}

, current through the resistance

R

would be

i_{2} = \frac{E _{1}}{R + I _{1}}

...(ii)
Since, from Eqs. (i) and (ii), we get

i_{2} > i_{1}

∴ \frac{E _{1}}{R + I _{1}} > \frac{E _{1} + E _{2}}{R + I _{1} + I _{2}}

This gives

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

Q. Under what conditions 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.

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

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

$E_{2} r_{2} < E (R + r_{2})$

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

Q. Under what conditions 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.

E2​r1​>E1​(R+r2​)

E1​r2​<E2​(r1​+R)

E2​r2​<E(R+r2​)

E1​r1​>E2​(r1​+R)

Q. Under what conditions 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.

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

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

$E_{2} r_{2} < E (R + r_{2})$

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