Question

In the network shown in figure, each resistance is $R$.The equivalent resistance between $A$ and $B$ is :

• A. $\frac{3R}{4}$
• B. $\frac{5R}{6}$
• C. $\frac{7R}{12}$
• D. $\frac{4R}{3}$

Answer 1

Answer: (C)

$\frac{7R}{12}$

If we connect a battery across A and B, then E and D will be at the same potential and F and C will be at the same potential.

Then the equivalent is as shown in figure.

here, $R_{eq1}=R||2R||R=(1/R+1/2R+1/R{)}^{-1}=(2R/5)$

Thus, equivalent resistance A and B is $R_{AB}=R||(R/2+2R/5+R/2)=R||\left(7R/5\right)=\frac{R\left(7R/5\right)}{R+7R/5}=\frac{7R}{12}$