A) a
B) b
C) c
D) d
Correct Answer: D
Solution :
Thermal energy in resistor is \[U={{i}^{2}}Rt\] where\[R={{R}_{0}}(1+\alpha t)\]Þ\[U={{i}^{2}}{{R}_{0}}(1+\alpha t)t={{i}^{2}}{{R}_{0}}t+{{i}^{2}}{{R}_{0}}{{t}^{2}}\] So \[\frac{dU}{dt}={{i}^{2}}{{R}_{0}}(1+\alpha t)\] With the time temperature increases, hence \[dU/dt\] increases. This is best shown by curve (d).
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