Question

Calculate the voltage $10V$ (in $V$ across the load resistor when two batteries of e.m.f $12V$ and $13V$ are connected in parallel across a load resistor of resistance $6\Omega$ . The internal resistances of the two batteries are $1\Omega$ and $2\Omega$ respectively.

Answer 1

Answer: 111

$E_1=12V{r}_1=1\Omega$

$E_{eq}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}=\frac{(12\times 2)+(13\times 1)}{1+2}=\frac{37}{3}V$
Also, $r_{eq}=\frac{r_1{r}_2}{r_1+r_2}=\frac{1\times 2}{1+2}=\frac{2}{3}\Omega$
Current in the circuit will be,
$I=\frac{E_{eq}}{R+r_{eq}}=\frac{\frac{37}{3}}{6+\frac{2}{3}}=\frac{37}{20}A$
The voltage across the load,
$V=IR=\frac{37}{20}\times 6=11.1V$
So, $10V=111V$