Question

Calculate the voltage $10\,V$ (in $V$ across the load resistor when two batteries of e.m.f $12\,V$ and $13\,V$ 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

$E_{1}=12\,V\,r_{1}=1\,\Omega$

$E_{eq}=\frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}}=\frac{\left(\right. 12 \times 2 \left.\right) + \left(\right. 13 \times 1 \left.\right)}{1 + 2}=\frac{37}{3}V$
Also, $r_{e q}=\frac{r_{1} r_{2}}{r_{1} + r_{2}}=\frac{1 \times 2}{1 + 2}=\frac{2}{3}\,\Omega$
Current in the circuit will be,
$I=\frac{E_{e q}}{R + r_{e q}}=\frac{\frac{37}{3}}{6 + \frac{2}{3}}=\frac{37}{20}A$
The voltage across the load,
$V=IR=\frac{37}{20}\times 6=11.1\,V$
So, $10\,V=111\,V$