It is desired to make a `20Omega` coil of wire, which has a zero thermal coefficient of resistance. To do this, a carbon resistor of resistance `R_1` is placed in series with an iron resistor of resistance `R_2`. The proportions of iron and carbon are so chosen that `R_1 +R_2 = 20Omega` for all temperatures near `20^@C.` How long are `R_1 and R_2`? `alpha_(carbon) =- 0.5xx10^(-3).^@C^(-1), alpha_(iron) = 5xx10^(-3) .^@C(-1)`

Explanation

We need `R_(1)(1+alpha_(1)trianglet)+R_(2)(1+alpha_(2)trianglet)=20`,because `R_(1)+R_(2)=20`where`trianglet=0`
We must have`R_(1)prop_(1)R_(2)prop_(2)whitprop_(1)=-0.5xx10^(-3)andprop_(2)=5xx10^(-3)`
Solving the two equations, `R_(1) + R_(2) = 20` and `R_(1 )= 18.18 Omega and R_(2) = 1.82 Omega`
(b) There is a symmetry about the line passing through QO and mid point of PR
`R_(PR)=(r)/(2)=2Omega`

( c) There is a symmetry about the line passing through E and mid point of CD.

`R_(AB)=(8)/(7)r=(8)/(7)Omega`
(d) The resistors `3Omega` and `6Omega` are in series and so are `5Omega` and `10Omega` resistors. These two series equivalents are in parallel to each other and also to the 4 Hence, the network reduces to the one given in the above figure`R_(eq)=5.34Omega`