Question

an infinite ladder with 1 ohm and 2 ohm resistors is constructed as shown in figure.Find the effective resistance between A and B with solution.answer fast please.

Answer 1

Answer:

This is a case of infinite series of resistances , so let's consider that the effective resistance be x

Now , since this is an infinite resistance series, if we skip the first square part, the other squares will form equivalent resistance of x (refer to attached photo to understand better)

So, x and 2 are in parallel and this combination is in series with 1 ohm.

$\dfrac{2x}{x + 2} + 1 = eq. \: r$

$= > \dfrac{2x}{x + 2} + 1 = x$

$= > \dfrac{2x + (x + 2)}{x + 2} = x$

$= > 3x + 2 = {x}^{2} +2 x$

$= > {x}^{2} - x - 2 = 0$

$= > {x}^{2} - 2x + x - 2 = 0$

$= > x(x - 2) + 1(x - 2) = 0$

$= > (x + 1)(x - 2) = 0$

So, either x = -1 or x = +2 . Since value of resistance has to be positive , we can say that Net Effective Resistance is 2 ohm.

So final answer :

$\boxed{ \red{ \bold{ \huge{eq. \: r \: = 2 ohm}}}}$

Answer 2

$\Large\boxed{\sf{\quad2\ \Omega\quad}}$

Consider the circuit with one $\sf{1\ \Omega}$ and $\sf{2\ \Omega}$ resistors each, connected as in the fig. Here the effective resistance is,

$\longrightarrow\sf{R(1)=(2+1)\ \Omega}$

$\longrightarrow\sf{R(1)=3\ \Omega}$

But we see that, here,

$\longrightarrow\sf{R(1)=\left(2+\dfrac{1}{1}\right)\ \Omega}$

Now consider the circuit with two $\sf{1\ \Omega}$ and $\sf{2\ \Omega}$ resistors each, connected as in the fig. Here the effective resistance is,

$\longrightarrow\sf{R(2)=\left(\left((2+1)^{-1}+2^{-1}\right)^{-1}+1\right)\ \Omega}$

$\longrightarrow\sf{R(2)=\dfrac{11}{5}\ \Omega}$

$\longrightarrow\sf{R(2)=\left(2+\dfrac{1}{5}\right)\ \Omega}$

$\longrightarrow\sf{R(2)=\left(2+\dfrac{1}{4\times1+1}\right)\ \Omega}$

Now consider the circuit with three $\sf{1\ \Omega}$ and $\sf{2\ \Omega}$ resistors each, connected as in the fig. Here the effective resistance is,

$\longrightarrow\sf{R(3)=\left(\left(\left(\left((2+1)^{-1}+2^{-1}\right)^{-1}+1\right)^{-1}+2^{-1}\right)^{-1}+1\right)\ \Omega}$

$\longrightarrow\sf{R(3)=\dfrac{43}{21}\ \Omega}$

$\longrightarrow\sf{R(3)=\left(2+\dfrac{1}{21}\right)\ \Omega}$

$\longrightarrow\sf{R(3)=\left(2+\dfrac{1}{4\times5+1}\right)\ \Omega}$

Now, consider the circuit with $\sf{n}$ no. of $\sf{1\ \Omega}$ and $\sf{2\ \Omega}$ resistors each, connected as in the fig. Here the effective resistance is,

$\longrightarrow\sf{R(n)=\left(2+\dfrac{1}{f(n)}\right)\ \Omega\quad\quad\dots(1)}$

where,

$\longrightarrow\sf{f(n)=4\cdot f(n-1)+1}$

And $\sf{f(1)=1}$ as seen earlier, so that $\sf{f(0)}$ should be $\sf{0.}$

$\longrightarrow\sf{f(1)=4\cdot f(0)+1}$     [By definition]

$\longrightarrow\sf{1=4\cdot f(0)+1}$

$\longrightarrow\sf{f(0)=0}$

Now let's have an expression for $\sf{f(n)}$ in terms of $\sf{n}$ only. By definition,

$\longrightarrow\sf{f(n)=4\cdot f(n-1)+1}$

$\longrightarrow\sf{f(n)=4[4\cdot f(n-2)+1]+1}$

$\longrightarrow\sf{f(n)=4^2\cdot f(n-2)+4+1}$

$\longrightarrow\sf{f(n)=4^2[4\cdot f(n-3)+1]+4+1}$

$\longrightarrow\sf{f(n)=4^3\cdot f(n-3)+4^2+4+1}$

$\sf{\dots\dots\dots\dots\dots}$

$\longrightarrow\sf{f(n)=4^n\cdot f(n-n)+4^{n-1}+4^{n-2}+\,\dots\,+4+1}$

Since $\sf{f(n-n)=f(0)=0,}$

$\longrightarrow\sf{f(n)=4^{n-1}+4^{n-2}+4^{n-3}+\,\dots\,+4+1}$

Well it's better to take,

$\longrightarrow\sf{f(n)=1+4+4^2+\,\dots\,+4^{n-1}}$

So $\sf{f(n)}$ appears as a geometric series with $\sf{a=1}$ and $\sf{r=4.}$

Hence,

$\longrightarrow\sf{f(n)=\dfrac{a(r^n-1)}{r-1}}$

$\longrightarrow\sf{f(n)=\dfrac{4^n-1}{3}}$

So (1) becomes,

$\longrightarrow\sf{R(n)=\left(2+\dfrac{3}{4^n-1}\right)\ \Omega}$

Hence the effective resistance in the infinite ladder shown in the fig. is given by,

$\displaystyle\longrightarrow\sf{R=\lim_{n\to\infty}\left(2+\dfrac{3}{4^n-1}\right)\ \Omega}$

Or,

$\displaystyle\longrightarrow\sf{R=\left(2+\lim_{n\to\infty}\dfrac{3}{4^n-1}\right)\ \Omega\quad\quad\dots(2)}$

Well,

• $\sf{n\longrightarrow\infty}$

• $\sf{4^n\longrightarrow\infty}$

• $\sf{4^n-1\longrightarrow\infty}$

• $\sf{\dfrac{1}{4^n-1}\longrightarrow0}$

• $\sf{\dfrac{3}{4^n-1}\longrightarrow0}$

Therefore,

• $\displaystyle\sf{\lim_{n\to\infty}\dfrac{3}{4^n-1}=0}$

So from (2),

$\longrightarrow\sf{R=(2+0)\ \Omega}$

$\longrightarrow\sf{\underline{\underline{R=2\ \Omega}}}$