Question

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 derives current in the circuit. What should be the value of r such that the bulb dissipated about 1 W of power?

• A. 14.8$\Omega$
• B. 29.6$\Omega$
• C. 7.4$\Omega$
• D. 3.7$\Omega$

Answer 1

The correct option is A

14.8$\Omega$

$P_{bulb}=\frac{v^2}{R}=i^{2}R$

I=$\frac{v^2}{16}$

$V_B$= 4 V

I =$i^2\times 16$

$I_B=\frac{1}{4}$ Amp.

6=$\frac{1}{4}\times r_{eq}$ (equivalent of groups of r)

where

$r_{eq}=r+\frac{r_{eq}.r}{r_{eq}+r}$