Question

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

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

Answer 1

The correct option is A $\frac{3R}{4}$

If we connect the battery across A and C, then no current will flow in the resistances between B and F and between H and D because the circuit is treated as balanced Wheatstone bridge. So they can be removed and corresponding equivalent circuit is shown in figure.

Here, equivalent resistance $R_{EG}=\frac{(R+R)(R+R)}{(R+R)+(R+R)}=R$

Now the equivalent resistance between A and C is:

$\frac{1}{R_{AC}}=\frac{1}{2R}+\frac{1}{3R}+\frac{1}{2R}=\frac{4}{3R}$

$\Rightarrow R_{AC}=\frac{3R}{4}$