Two batteries of emf $E_{1}$ and $E_{2} (E_{2} > E_{1})$ and internal resistance $r_{1}$ and $r_{2}$ respectively are connected in parallel as shown in figure.

Open in App

Solution

We have two cells $E_{1}$ and $E_{2}$ with internal resistance $r_{1}$ and $r_{2}$ connected in parallel with a resistor $R$ .
Let $I_{1}$ and $I_{2}$ be the current through $E_{1}$ and $E_{2}$ respectively.
By applying Kirchhoff's current law
$I = I_{1} + I_{2}$
Let there resultant EMF is $E_{e f f}$ and resultant resistance be $R_{e q}$
By Kirchhoff's voltage flow
Potential across $R = E_{e f f} - I R_{e q}$
Potential across upper branch $= E_{1} - I r_{1}$
Potential across upper branch $= E_{2} - I r_{2}$
After solving we will get
$E_{e f f} = (\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}) R_{e q}$
And
$R_{e q} = \frac{r_{1} r_{2}}{(r_{1} + r_{2})}$

Suggest Corrections

Similar questions

Q. Two batteries of emf

ϵ_{1}

and

ϵ_{2} (ϵ_{2} > ϵ_{1})

and internal resistances

r_{1}

and

r_{2}

respectively are connected in parallel as shown in figure.

A single battery fo emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs $E_{1} = 2 V$ and $E_{2} = 1.5 V$ and internal resistances $r_{1} = 1 Ω$ and $r_{2} = 2 Ω$ respectively, with polarities as shown in the figure.

(IIT-JEE 1997)