# A bulb of resistance R=16 ohms is attached in series with an infinite resistor network with identical resistances r ohms. A 10 V battery drives current in the circuit. What should be the value of 'r' such that the bulb dissipates about 1 W of power?

## Thank you!

Jan 5, 2017

$14.8 \Omega$, rounded to one decimal place.

#### Explanation:

Let ${R}_{e}$ be equivalent resistance of an infinite resistor network with identical resistances of $r$ ohms. Since bulb is connected in series with network,

Total resistance seen by battery $= R + {R}_{e}$
Current in the bulb is given by the expression
$I = \frac{V}{R + {R}_{e}}$ ......(1)
Power dissipated in the bulb is
$P = {I}^{2} R$ .....(2)
Inserting given values we get from (1)
$I = \frac{10}{16 + {R}_{e}}$ and inserting this value and other given numbers we get from (2)
$1 = {\left(\frac{10}{16 + {R}_{e}}\right)}^{2} \times 16$
Taking square root of both sides we get
$1 = \frac{10}{16 + {R}_{e}} \times 4$
$\implies 16 + {R}_{e} = 40$
$\implies {R}_{e} = 40 - 16 = 24 \Omega$ ......(3)

To calculate ${R}_{e}$
As there are infinite many resistors, there will still be infinite many resistors if we detach the first two resistors from the front of nodes $X \mathmr{and} Y$ as shown in figure. Resistance seen looking to the right will be the same as the resistance seen between nodes $A \mathmr{and} B$

As such the network reduces to sum of two resistances 1. resistance $r$ between nodes $A \mathmr{and} B$ and 2. resistance equivalent to two parallel resistors $r$ and ${R}_{e}$ between nodes $X \mathmr{and} Y$.

For infinite resistor network we have an equation

${R}_{e} = r + \frac{r \times {R}_{e}}{r + {R}_{e}}$
$\implies \left({R}_{e} - r\right) = \frac{r \times {R}_{e}}{r + {R}_{e}}$
$\implies \left({R}_{e} - r\right) \times \left(r + {R}_{e}\right) = \left(r \times {R}_{e}\right)$
$\implies \left({R}_{e}^{2} - {r}^{2}\right) - \left(r \times {R}_{e}\right) = 0$

Using (3) we get

$\left({24}^{2} - {r}^{2}\right) - 24 r = 0$
$\implies {r}^{2} + 24 r - {24}^{2} = 0$

Solving the quadratic and choosing positive root as resistance can not be negative
$r = \frac{- 24 \pm \sqrt{{24}^{2} + 4 \times {24}^{2}}}{2}$
$r = 14.8 \Omega$, rounded to one decimal place.

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Quadratic equation can also be solved using inbuilt graphic utility