For the circuit shown in Figure, the equivalent resistance between $A$ and $C$ is

\frac{12}{11} r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{13}{11} r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{14}{11} r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{15}{11} r

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{15}{11} r$
By Kirchhoff's current law
at junction E : $I_{2} - I_{3} = I_{1} + I_{3} + I_{4} \Rightarrow I_{1} - I_{2} = - 2 I_{3} - I_{4} . . . . (1)$
By Kirchhoff's voltage law
for loop AFBA: $r I_{1} = r I_{2} + r I_{3} \Rightarrow I_{1} - I_{2} = I_{3} . . . . . (2)$
for loop BEFB: $r (I_{2} - I_{3}) + r I_{4} = r I_{3} \Rightarrow I_{4} = 2 I_{3} - I_{2} . . . . . (3)$
for loop AFDCMA: $V = r I_{1} + r (I_{2} - I_{3}) + r I_{2} = r I_{1} + 2 r I_{2} - r I_{3} \Rightarrow I_{1} + 2 I_{2} - I_{3} = V / r . . . . (4)$
Eliminating $I_{3}, I_{4}$ from (1) using (2) and (3), $I_{1} - I_{2} = - 2 (I_{1} - I_{2}) + I_{2} - 2 I_{3}$
or $I_{1} - I_{2} = - 2 I_{1} + 2 I_{2} + I_{2} - 2 (I_{1} - I_{2})$
or $5 I_{1} = 6 I_{2} \Rightarrow I_{1} = (6 / 5) I_{2}$
from (2) and (4), $I_{1} + 2 I_{2} - I_{1} + I_{2} = V / r \Rightarrow I_{2} = V / 3 r$
thus, $I_{1} = (6 / 5) (V / 3 r) = 2 V / 5 r$
The equivalent resistance $R_{A C} = \frac{V}{I} = \frac{V}{I_{1} + I_{2}} = \frac{V}{2 V / 5 r + V / 3 r} = \frac{15 r}{11}$

Suggest Corrections

Similar questions

\infty \sum r = 1 (2 r - 1) {(\frac{9}{11})}^{r}

\frac{11 r}{4} - 21 = 12

H e

H_{2}

Join BYJU'S Learning Program