Question

The value of effective resistance between $A$ and $B$ is? $\left( {Given\,\,R = 2k\Omega } \right)$

$\left( a \right)\,\,(1 + \sqrt 3 )k\Omega$
$\left( b \right)\,\,2(1 + \sqrt 3 )k\Omega$
$\left( c \right)\,\,(1 - \sqrt 3 )k\Omega$
$\left( d \right)\,\,2(1 - \sqrt 3 )k\Omega$

Answer 1

Whenever you are given with a question where resistances or capacitances are connected in a similar pattern up to $'n'$ number of times, firstly find unit cell of the network then assume x or y as the combined resistance for the rest of pattern $\left( {n - 1} \right)$ number of times. This pattern is also known as the infinite ladder network.
Formula used:
(A) $r = {r_1} + {r_2} + {r_3} + ........... + {r_n}$ (when resistance are connected in series)
(B) $\dfrac{1}{r} = \dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}} + ............. + \dfrac{1}{{{r_n}}}$ (when resistance are connected in parallel)

Complete step by step answer:
Let us assume that equivalent resistance or effective resistance for this network is ${R_{eq}}$ . Here, in this problem, the unit cell of network is shown in below picture as follows:

Thus, your first basic step in the question is to find a unit cell of the network in circuit.
Let the combined resistance for the rest of the network (except the resistance of unit cell 1) be $x$ .
$\Rightarrow {R_{AB}} = x$
Where, ${R_{AB}}$ is equal to the combined resistance of $\left( {n - 1} \right)$ unit cells.

Then,

Also,

From the figure, the combined resistance $r$ for parallel combination is given by
$\dfrac{1}{r} = \dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}} + ............. + \dfrac{1}{{{r_n}}}$ (When resistance are connected in parallel)
And it can be written as
$\Rightarrow \dfrac{1}{r} = \dfrac{1}{R} + \dfrac{1}{x}$
Now on solving it, we get
$\Rightarrow \dfrac{1}{r} = \dfrac{{x + R}}{{xR}}$
Solving it, we get
$\Rightarrow r = \dfrac{{xR}}{{x + R}}$ and we will let it eq. $1$

Next, from the below shown figure, it’s clear that all the three resistances are in connected in series,

$\therefore$ By applying, $r = {r_1} + {r_2} + {r_3} + ........... + {r_n}$ (when resistance are connected in series) we get,${R_{eq}} = R + r + R$ and we will let it eq. $2$
Also, given in question that, $R = 2K\Omega$ and from above eq. $1$ , substituting values in eq. $2$ , we get as follows:
$\Rightarrow {R_{eq}} = 2 + \dfrac{{2x}}{{2 + x}} + 2$
And on solving it, we get
$\Rightarrow {R_{eq}} = 4 + \dfrac{{2x}}{{2 + x}}$
Now, ${R_{eq}} \approx {R_{AB}}$
$\Rightarrow {R_{eq}} = x$ , thus
$\Rightarrow x = 4 + \dfrac{{2x}}{{2 + x}}$
Now on doing the cross multiplication, we get
$\Rightarrow x\left( {2 + x} \right) = 4\left( {2 + x} \right) + 2x$
And on solving it, we get
$\Rightarrow 2x + {x^2} = 8 + 4x + 2x$
Further doing the solution,
$\Rightarrow {x^2} = 8 + 4x$
Or it can be written as
$\Rightarrow {x^2} - 4x - 8 = 0$
By applying the formula for finding roots of a quadratic equation, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , We get,
$\Rightarrow \dfrac{{4 \pm \sqrt {16 + 32} }}{2}$
On solving it,
$\Rightarrow \dfrac{{4 \pm \sqrt {48} }}{2}$
And it will be equal to
$\Rightarrow 2(1 \pm \sqrt 3 )$
Now the value of $\sqrt 3 = 1.7320$, so $1 - \sqrt 3$ have negative value, and resistance can’t have a negative value.
$\therefore x = 2(1 + \sqrt 3 )$ would be the effective resistance to give an infinite ladder network.

$\Rightarrow$ Option $\left( b \right)$ is correct.

Note: Do not confuse yourself with why we put${R_{eq}} \approx {R_{AB}}$ , because ladder continues up to infinity, therefore, resistance of $1unit$ cell would be very-very small and can be negligible. Always, consider the positive value of resistance instead a negative value. Try to simplify the network step by step to reduce any error.