In the figure,the value of the resistor to be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the equivalent resistance of the entire infinite ladder network between $A$ and $B$ be $R_0$. If we add one more elementary set to the network

Vedclass · Accepted Answer

$R(\sqrt{3} - 1)$

Vedclass · Answer

(B) Let the equivalent resistance of the entire infinite ladder network between $A$ and $B$ be $R_0$. If we add one more elementary set to the network,the total resistance remains $R_0$. Consider the first elementary set consisting of two resistors $R$ in series with the upper and lower arms,and one resistor $R$ in parallel. The remaining infinite network to the right of this set also has an equivalent resistance $R_0$. The total resistance $R_{AB}$ is given by the parallel combination of the vertical resistor $R$ and the series combination of the upper arm resistor $R$,the right-side equivalent resistance $R_0$,and the lower arm resistor $R$. Thus,$R_{AB} = \frac{R(R + R_0 + R)}{R + (R + R_0 + R)} = R_0$. $R(2R + R_0) = R_0(3R + R_0)$ $2R^2 + RR_0 = 3RR_0 + R_0^2$ $R_0^2 + 2RR_0 - 2R^2 = 0$. Solving this quadratic equation for $R_0$ using the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $R_0 = \frac{-2R \pm \sqrt{(2R)^2 - 4(1)(-2R^2)}}{2} = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2} = \frac{-2R \pm \sqrt{12R^2}}{2} = \frac{-2R \pm 2R\sqrt{3}}{2} = R(\sqrt{3} - 1)$ (taking the positive value).

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

Similar Questions

$A$ $5\ V$ battery with internal resistance $2\,\Omega$ and a $2\,V$ battery with internal resistance $1\,\Omega$ are connected to a $10\,\Omega$ resistor as shown in the figure. Find the current through the $10\,\Omega$ resistor.

In the circuit shown,the variable resistance is adjusted such that the ammeter reading is the same in both positions $1$ and $2$ of the key. The reading of the ammeter is $2 \, A$. If $E = 20 \, V$,then $x$ is: ................... $\Omega$

In the circuit shown,wire $1$ is of negligible resistance. Then,

$A$ light bulb of resistance $R = 16 \, \Omega$ is attached in series with an infinite resistor network with identical resistances $r$ as shown below. $A$ $10 \, V$ battery drives current in the circuit. Find the value of $r$ in $\Omega$ such that the bulb dissipates about $1 \, W$ of power.

Vedclass Test Series

Exam Paper Generator

Online Exam Module