Question

In the circuit shown find the equivalent resistance between $A$ and $B$.

Answer 1

From the above figure,

$R_{eq}=\frac{E}{I}$

loop $CFAC:E=r{I}_1+R(I_1-I_2)\Rightarrow E=(r+R)I_1-R{I}_2$

loop $ACBDEA:r{I}_1+r{I}_2=R(I-I_1)\Rightarrow (r+P)I_1+r{I}_2=RI$

$r{I}_2+r(I-I_1+I_2)=R(I_1-I_2)$

$\Rightarrow (r+R)I_3-(R+2r)I_2=rI$

From Eqs (ii)-(iii), we get $(3r+R)I_2=(R-r)I\Rightarrow I_2=\frac{(R-r)I}{3r+R}$

Putting the above value in eq (ii), we get

$(r+R)I_1+\frac{r(R-r)}{3r+R}I=RI\Rightarrow I_1=\frac{(R-r)I}{3r+R}$

Putting $I_1$ and $I_2$ in Eq (i) we get

$R_{eq}=\frac{E}{I}=\frac{(R+r{)}^2}{3r+R}-\frac{r(R-r)}{3r+R}=\frac{r(r+3R)}{3r+R}$