A straight nicrome wire is initially at room temperature `20^(@)C` it is connected to an ideal battery of 500 volt. Just after switching on, the current detected is 5 amp. Due to heaing effect its temperature increases. and it also loosing heat to the environment according to newton's cooling law as `(dQ_(max))/(dt)=45(T-20^(@))J//sec`. at steady state the current detected is 4.5 amp.

Options

A. Steady state temperature of the wire is `70^(@)C`
B. steady state temperature of the wire is `75.5^(@)C`
C. temperature co-efficient of resistance of the wire is nearly `2.2xx10^(-3)//.^(@)C`
D. temperature co-efficient of resistance of the wire is nearly `1.57.10^(-3)//.^(@)C`

Explanation

For steady state
`((dQ)/(dt))_("in")=((dQ)/(dt))_(out)`
`(V)(i_(55))=45(T-20)`
`(500)(4.5)=45(T-20)`
`T_(55)=70^(@)C`
Resistance at `20^(@)C` is `R=(v)/(i)=(500)/(50)`
`R_(20)=100Omega`
Resistance at `70^(@)C` is `R=(v)/(i)=(500)/(4.5)~=111Omega`
`R_(f)=R_(0)(1+alphatriangleT)`
`111=1000(1+alpha(50))`
`alpha=(0.11)/(50)=2.2xx10^(-3)//.^(@)C`.