A potential difference of 120 V is applied to a coil at temperature of `12^@C` and the current is 6A. What will be the mean temperature of the coil when the curreny has fallen to 3 A, the applied voltage being the same as before ? Given temperature coefficient of resistance coil is `.00427^@C^(-1)` at `0^@C`.

Explanation

In second case the current decreases due to incerese in resistance on heating. Here
`R_(12)=V/I=120/6=20 Omega`
If `t^@C` is the temperature at which current falls
to 3A, then `= R_(1)=120/3=40 Omega`
As `R_(t)=R_(0)(1+alpha t)`
So `R_(12)=R_(0)(1+alpha xx12)`
or `20 =R_(0)(1+.00427 xx 12)`
and `R_(t) = R_(0)(1+alpha t)or 40=R_(0)(1+.00427 t)`
`:. 40/20 =(1+0.00427 t)/(1+0.00727 xx12)`
On solving, `t=258^@C`.