Thank you for reporting, we will resolve it shortly
Q.
For the circuit shown in figure, the equivalent resistance between $A$ and $C$ is
Current Electricity
Solution:
At junction $E$ or $F$ :
$I_{1}+I_{3}+I_{4}=I_{2}-I_{3}$
or $I_{1}-I_{2}+2 I_{3}+I_{4}=0$
Loop $A F B A$ or loop $E C D E$ :
$r I_{1}=r I_{2}+r I_{3}$
or $I_{1}=I_{2}+I_{3}$
Loop $B E F B$ or loop $E D F E$ :
$r\left(I_{2}-I_{3}\right)+r I_{4}=r I_{3} \text { or } I_{2}-2 I_{3}+I_{4}=0$
Loop $A F D C M A$ or $A B E C M A$ :
$V=r I_{1}+r\left(I_{2}-I_{3}\right)+r I_{2}=r I_{1}+2 r I_{2}-r I_{3}$
Solve to get $I_{1}=2 V / 5 r, I_{2}=V / 3 r$
$R_{ eq }=\frac{V}{I}=\frac{V}{I_{1}+I_{2}}$
$=\frac{V}{2 V / 5 r+V / 3 r}=\frac{15 r}{11}$
