A light bulb of resistance R = 16 , Omega is | vedclass

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(A) Let the equivalent resistance of the infinite network be $x$. Adding or removing one repeating unit does not alter the resistance of an infinite n

Vedclass · Accepted Answer

$14.8$

Vedclass · Answer

(A) Let the equivalent resistance of the infinite network be $x$. Adding or removing one repeating unit does not alter the resistance of an infinite network. Thus, the network can be represented as a series combination of $r$ and a parallel combination of $r$ and $x$.$x = r + \frac{rx}{r+x}$$x(r+x) = r(r+x) + rx$$xr + x^2 = r^2 + rx + rx$$x^2 - rx - r^2 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{r \pm \sqrt{r^2 - 4(1)(-r^2)}}{2} = \frac{r \pm \sqrt{5r^2}}{2} = \frac{r(1 + \sqrt{5})}{2}$ (taking the positive root).The power consumed by the bulb is $P = i^2 R = 1 \, W$.$i^2 (16) = 1 \Rightarrow i^2 = \frac{1}{16} \Rightarrow i = 0.25 \, A$.The total resistance of the circuit is $R_{total} = R + x = 16 + \frac{r(1 + \sqrt{5})}{2}$.Using Ohm's law $V = i R_{total}$:$10 = 0.25 	imes (16 + \frac{r(1 + \sqrt{5})}{2})$$40 = 16 + \frac{r(1 + 2.236)}{2}$$24 = \frac{3.236r}{2} = 1.618r$$r = \frac{24}{1.618} \approx 14.83 \, \Omega$.Thus, the value of $r$ is approximately $14.8 \, \Omega$.

$A$ 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.

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