Question

Find the equivalent resistance between A and E (Resistance of each resistor is R).

• A. $\frac{7}{12}R$
• B. $\frac{7}{13}R$
• C. $\frac{7}{15}R$
• D. $\frac{7}{13}R$

Answer 1

The correct option is A $\frac{7}{12}R$

From A's perspective, point B and D should be at same potential since both of them are connected to A with single resistor of equal resistance. Also point B,I and D,J are connected by wire of negligible resistance, therefore I and J will be at same potential as are points B and D. Thus point B, D, I and J is at same potential.
From E's perspective, F and H should be at same potential since both are connected by single resistor of equal resistance.
Therefore equivalent circuit for the above can be drawn as shown below.
Two resistors of resistance $R$ in parallel will have equivalent resistance $R/2$
Hence the circuit reduces to
The lower branch BCGF having resistors $R/2$, $R$ and $R/2$ in series will have equivalent resistance, $R_{eq1}=R/2+R+R/2=2R$
This equivalent resistor of resistance $2R$ is connected in parallel to resistor $R/2$ across BF. Therefore equivalent resistance will be $R_{eq2}=\frac{2R\times \left(R/2\right)}{2R+\left(R/2\right)}=\frac{2R}{5}$
Now circuit gets reduced to one as shown below.
Since $R/2$, $2R/5$ and $R/2$ are in series, their equivalent will be $R/2+2R/5+R/2=7R/5$.
This equivalent resistance $7R/5$ is in parallel with resistor $R$, therefore equivalent resistance across A and E will be,
$R_{eq}=\frac{\left(7R/5\right)\times R}{\left(7R/5\right)+R}=\frac{7R}{12}$
Hence option A is correct.