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 $10V$ battery derives current in the circuit. What should be the value of $r$ such that the bulb dissipated about $1W$ 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 $14.8\Omega$

$P_{bulb}=i^2{R}_{bulb}....\left(1\right)$
$i=\frac{v}{R_{total}}....\left(2\right)$
Now $R_{total}=R_{bulb}+r_{eq}....\left(3\right)$
Where $r_{eq}$ is the equivalent resistance of the infinite circuit.

Now the expression for $r_{eq}$ can be given by (refer the figure)
$r_{eq}=r+\frac{r_{eq}r}{r_{eq}+r}$

Solving the above quadratic equation we get the value of $r_{eq}$ as $\frac{1+\sqrt{5}}{2}r$

Putting this value of $R_{eq}$ in equation $\left(3\right)$ and put $R_{bulb}=16\Omega$

Put this value of $R_{total}$ in equation $\left(2\right)$ and put the expression of $i$ in equation $\left(1\right)$ We get the value of r as $14.8\Omega$