Question

Find the value of $i_1/i_2$ in figure if (a) $R=0.1\Omega$, (b) $R=1\Omega$ (c) $R=10\Omega$. Note from your answers that in order to get more current from a combination of two batteries they should be joined in parallel if the external resistance is small and in series if the external resistance is small and in series if the external resistance is large as compared to the internal resistances.

Answer 1

$a)$ $0.1i_1+1i_1-6+1i_1-6=0$

$\Rightarrow 0.1i_1+1i_1+1i_1=12$

$\Rightarrow i_1=\frac{12}{2.1}=5.71A$

$ABCDA$

$\Rightarrow 0.1i_2+1i-6=0$

$\Rightarrow 0.1i_2+i=6$

$\Rightarrow i=6-0.1i_2$

$ADEFA$,

$\Rightarrow i-6+6-(i_2-i)1=0$

$\Rightarrow i-i_2+i=0$

$\Rightarrow 2i-i_2=0$

$\Rightarrow 2[6-0.1i_2]-i_2=0$

$\Rightarrow i_2=10A$.

$\therefore \frac{i_1}{i_2}=0.571A$

$b)$ $1i_1+1i_1-6+1i_1=0$

$\Rightarrow 3i_1=12\Rightarrow i_1=4$

$ABCDA$,

$i_2+(i_2-i)-6=0$

$\Rightarrow i_2+i_2-i=6\Rightarrow 2i_2-i=6$

$\Rightarrow -2i_2\pm 2i=6\Rightarrow i=-2$

$i_2+i=6$

$ADEFA$

$i-6+6-(i_2-i)=0$

$\Rightarrow i-i_2+i=0$

$\Rightarrow 2i-i_2=0$

$\Rightarrow 2[6-i_2]-i_2=0$

$\Rightarrow 12-3i_2=0$

$\Rightarrow i_2=4A$

$\Rightarrow \frac{i_1}{i_2}=1$

$c)$ $10i_1+1i_1-6+1i_1-6=0$

$\Rightarrow 12i_1=12\Rightarrow i_1=1$

$ABCDA$

$10i_2-i_1-6=0$

$\Rightarrow i=6-10i_2$

$ADEFA$

$i-6+6-(i_2-i)1=0$

$\Rightarrow i-i_2+i=0$

$\Rightarrow 2i-i_2=0$

$\Rightarrow 2[6-10i_2]-i_2=0$

$\Rightarrow 12-21i_2=0$

$i_2=0.57A$

$\therefore \frac{i_1}{i_2}=1.75$