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!

1 Answer
Jan 5, 2017

Answer:

#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,

www.doorsteptutor.com

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.

-.-.-.-.-.-.-.

Quadratic equation can also be solved using inbuilt graphic utility
inbuilt graphic utility