The resistance of a conductor is 5 ohm at 50*c and 6 ohm at 100*c.Its resistance at 0*is??THANK YOU!!

2 Answers
May 30, 2015

Well, try thinking about it this way: the resistance changed by only #1 Omega# over #50^oC#, which is a pretty large temperature range. So, I would say it's safe to assume the change in resistance with respect to temperature (#(DeltaOmega)/(DeltaT)#) is pretty much linear.

#(DeltaOmega)/(DeltaT) ~~ (1 Omega)/(50^oC)#

#DeltaOmega = (1 Omega)/(100^oC-50^oC)*(0^oC-50^oC) ~~ -1 Omega#

#Omega_(0^oC) ~~ 4 Omega#

Its resistance at #0^@"C""# is 4 ohm.

Explanation:

#R_T=(1+alpha T)R#, where

#R_T=#Resistance at any temperature,
#alpha#=constant of material,
#R=#resistance at Zero degree Celsius.

At 50 degrees Celsius:

#R_50=(1+50alpha )R#=#"5 ohm"# #" "color(blue)((1))#

At 100 degrees Celsius:

#R_100=(1+100alpha)R = "6 ohm"# #" "color(blue)((2))#

At zero degrees Celsius:

#R_0=(1+0)R#
#R_0=R# #" "color(blue)((3))#

Determination R from equations #color(blue)((1))# and #color(blue)((2))#** by

#color(blue)((1))/color(blue)((2)) => (1+50alpha)/(1+100alpha) =5/6#

#6 + 300alpha = 5 + 500alpha => alpha = 1/200#

Use this value in equation #color(blue)((1))#

#(1+ 1/200 * 50) * R = 5 => 5/4* R = 5 => R = "4 ohm"#

According to equation #color(blue)((3))#, you have

#R_0 = R = "4 ohm"#

Therefore, its resistance at #0^@"C"# is #"4 ohm"#.