Question

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!

Answer 1

`14.8Omega`, 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=V/(R+R_e)` ......(1)
Power dissipated in the bulb is
`P=I^2R` .....(2)
Inserting given values we get from (1)
`I=10/(16+R_e)` and inserting this value and other given numbers we get from (2)
`1=(10/(16+R_e))^2xx16`
Taking square root of both sides we get
`1=10/(16+R_e)xx4`
`=>16+R_e=40`
`=>R_e=40-16=24Omega` ......(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 and Y` as shown in figure. Resistance seen looking to the right will be the same as the resistance seen between nodes `A and B`

As such the network reduces to sum of two resistances 1. resistance `r` between nodes `A and B` and 2. resistance equivalent to two parallel resistors `r` and `R_e` between nodes `X and Y`.

For infinite resistor network we have an equation

`R_e=r+(rxxR_e)/(r+R_e)`
`=>(R_e-r)=(rxxR_e)/(r+R_e)`
`=>(R_e-r)xx(r+R_e)=(rxxR_e)`
`=>(R_e^2-r^2)-(rxxR_e)=0`

Using (3) we get

`(24^2-r^2)-24r=0`
`=>r^2 +24r-24^2=0`

Solving the quadratic and choosing positive root as resistance can not be negative
`r=(-24+-sqrt(24^2+4xx24^2))/2`
`r=14.8Omega`, rounded to one decimal place.

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Quadratic equation can also be solved using inbuilt graphic utility