Question

In figure, find the value of the resistor to be connected between C and D, so that the resistance of the entire circuit between A and B does not change with the number of elementary sets.

$\begin{align}
  & \left( A \right)R \\
 & \left( B \right)R\left( \sqrt{3}-1 \right) \\
 & \left( C \right)3R \\
 & \left( D \right)R\left( \sqrt{3}+1 \right) \\
\end{align}$

Answer 1

Hint: Consider the new resistance across CD as ‘x’. Then the equivalent resistance across CD will be the same as the resistance across AB and EF. When we take the resistance across CD as ‘x’, then resistance across AB and EF will also be ‘x’. Then draw the equivalent circuit and then find the resistance across AB. We have already considered that resistance across AB is x. Thus equate the both terms and using quadratic equation find the value of ‘x’ and that will be resistance across CD.

Complete answer:
Consider the new resistance across the CD as ‘x’. Then the equivalent resistance across CD will be the same as the resistance across AB and EF. When we take the resistance across CD as ‘x’, then resistance across AB and EF will also be ‘x’.
Thus let resistance across AB and EF be ‘x’. For this consider a part of the circuit as shown below.

Then the equivalent circuit can be drawn as,

That is, here the three resistors R, x and R are connected in series.
Hence there equivalent resistance is,
$\begin{align}
  & {{R}_{eq}}=R+x+R \\
 & \Rightarrow {{R}_{eq}}=2R+x \\
\end{align}$
These two resistors are connected in parallel. Hence, resistance across AB is,
$\dfrac{1}{{{R}_{AB}}}=\dfrac{1}{R}+\dfrac{1}{2R+x}$
$\begin{align}
  & \Rightarrow \dfrac{1}{{{R}_{AB}}}=\dfrac{2R+x+R}{R\left( 2R+x \right)} \\
 & \Rightarrow {{R}_{AB}}=\dfrac{R\left( 2R+x \right)}{3R+x} \\
\end{align}$
$\Rightarrow {{R}_{AB}}=x$
Thus,
$\dfrac{R\left( 2R+x \right)}{3R+x}=x$
$R\left( 2R+x \right)=3Rx+{{x}^{2}}$
$2{{R}^{2}}+Rx=3Rx+{{x}^{2}}$
$\Rightarrow {{x}^{2}}+2Rx-2{{R}^{2}}=0$
Apply quadratic equations.
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$\Rightarrow x=\dfrac{-2R\pm \sqrt{{{\left( 2R \right)}^{2}}-4\times 1\times \left( -2{{R}^{2}} \right)}}{2}$
Then,
$x=\dfrac{-2R\pm \sqrt{4{{R}^{2}}+8{{R}^{2}}}}{2}$
$\Rightarrow x=\dfrac{-2R\pm \sqrt{12{{R}^{2}}}}{2}$
\[\begin{align}
  & \Rightarrow x=\dfrac{-2R\pm 2R\times \sqrt{3}}{2} \\
 & \Rightarrow x=-R\pm R\sqrt{3} \\
\end{align}\]
$\therefore x=R\left( -1\pm \sqrt{3} \right)$
Resistance should be a positive value.
Hence, ${{R}_{CD}}=x=\left( \sqrt{3}-1 \right)R$

Note:
The net resistance in a series connection is the sum of resistances connected in the given circuit. Whereas, the net resistance in a parallel connection, that is the reciprocal of equivalent resistance is the sum of reciprocals of the individual resistances. Hence while calculating the equivalent resistance, first note whether it is connected in series or parallel. Then use the suitable equation of the given circuit.