Question

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.

• A. $E_{2} r_{1}>E_{1}\left(R+r_{2}\right)$
• B. $E_{1} r_{2}< E_{2}\left(r_{1}+R\right)$
• C. $E_{2} r_{2}< E\left(R+r_{2}\right)$
• D. $E_{1} r_{1}>E_{2}\left(r_{1}+R\right)$

Answer 1

Answer: (B)

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} $
$ \therefore \frac{E_{1}}{R+I_{1}} >\frac{E_{1}+E_{2}}{R+I_{1}+I_{2}}$
This gives $E_{1} I_{2}>E_{2}\left(R+r_{1}\right)$