Find the emf(V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs $V_{1}$ and $V_{2}$ and internal resistance $r_{1}$ and $r_{2}$ respectively, with polarities as shown in the figure.

Open in App

Solution

The equivalent resistance, $r_{e q} = \frac{r_{1} r_{2}}{r_{1} + r_{2}}$

This is the internal resistance of the single battery which is
equivalent to a parallel combination of the two batteries. Now
applying Kirchhoff s law in loop HCDEFGH moving clockwise
starting from D
$- I r_{2} + V_{2} + V_{1} - I r_{1} = 0$
$I = \frac{V_{1} + V_{2}}{r_{1} + r_{2}}$
$V_{A} + I r_{1} - V_{1} = V_{B}$
$V_{A} - V_{B} = V_{1} - I r_{1}$
$V_{A} - V_{B} = V_{1} - \frac{(V_{1} + V_{2}) r_{1}}{r_{1} + r_{2}} = \frac{V_{1} r_{1} + V_{1} r_{2} - V_{1} r_{1} - V_{2} r_{1}}{r_{1} + r_{2}}$

$V_{A} - V_{B} = \frac{V_{1} r_{2} - V_{2} r_{1}}{r_{i} + r_{2}}$
Hence, emf is $\frac{V_{1} r_{2} - V_{2} r_{1}}{r_{i} + r_{2}}$ . and equivalent resistance is $r_{e q} = \frac{r_{1} r_{2}}{r_{1} + r_{2}}$

Suggest Corrections

Similar questions

Q. The emf (V) and internal resistance (

r

) of a single battery which is equivalent to a parallel combination of two batteries of emf's

V_{1}

and

V_{2}

and internal resistance

r_{1}

and

r_{2}

, respectively, with polarities as shown in the figure are given by

\frac{V_{1} r_{2} - V_{2} r_{1}}{r_{1} + r_{2}}, \frac{r_{1} r_{2}}{r_{1} + r_{2}}

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)