Question

Find the effective resistance between $\text{A}$ & $\text{B}$.

• A. $\frac{7R}{12}$
• B. $\frac{R(1+\sqrt{3})}{2}$
• C. $\frac{R(1+\sqrt{5})}{2}$
• D. $\frac{R}{12}$

Answer 1

The correct option is C $\frac{R(1+\sqrt{5})}{2}$

Let the effective resistance between $\text{A}$ & $\text{B}$ be $R_E$.

Since the network is infinite long, so removal of one group of resistance from the chain will not change the network.

Therefore, effective resistance between points $\text{C}$ & $\text{D}$ would also be $R_E$.

The equivalent network will be as shown below.


The original infinite chain is equivalent to $R$ in series with $R$ & $R_E$ in parallel. So the effective resistance of infinite network will be

$R_E=R+\frac{R{R}_E}{R+R_E}$

$\Rightarrow R_{E}R+R_E^2=R^2+2R{R}_E$

$\Rightarrow R_E^2-R{R}_E-R^2=0$

$\therefore R_E=\frac{R\pm \sqrt{R^2-4\cdot 1\cdot (-R^2)}}{2\cdot 1}$

$\Rightarrow R_E=\frac{R(1\pm \sqrt{5})}{2}$

Taking positive value, because resistance can't be negative.

$\therefore R_E=\frac{R(1+\sqrt{5})}{2}$

Hence, option $\left(a\right)$ is correct.

Key Concept: In an infinitely long network, addition or removal of one section from the chain still leaves the chain infinite (i.e. just about the same as before).