Question

A potential difference of 250V is applied to a field winding at 15°C and the current is 5A. What will the temperature of the winding when current has fallen to 3.91A, applied voltage being constant. Assume delta15=0.00393/°C

Answer 1

$$R(15°C)=\frac{V}{I_1}=\frac{250}{5}=50\:\Omega$$

$$R(t)=\frac{V}{I_2}=\frac{250}{3.91}=63.9\:\Omega$$

$$R(t)=R_0(1+\alpha t)\\ R(15{^\circ C})=R_0(1+\alpha 15)$$

$$t=\frac{1}{\alpha}(R(t)/R(15{^\circ C})(1+15\alpha)-1)$$

$$t=\frac{1}{0.00393}(63.9/50(1+15*0.00393)-1)=90{^\circ C}$$