Question

Two batteries with e.m.f. $12\, V$ and $13\, V$ are connected in parallel across a load resistor of $6 \,\Omega$. The internal resistances of the two batteries are $1 \,\Omega$ and $2 \,\Omega$ respectively. The voltage across the load is ______ volts

Answer 1

$E_{e q}=\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1}+r_{2}}$
$=\frac{(12 \times 2)+(13 \times 1)}{1+2}=\frac{37}{3} V$
Also, $r_{e q}=\frac{r_{1} r_{2}}{r_{1}+r_{2}}$
$=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$
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,
$V = IR =\frac{37}{20} \times 6=11.1\, V$