A.The amount of heat generated in a power supply can be expressed by its internal loss as follows:

Amount of heat generated: 1(W x s) = 1(J) = 4.2(cal)

The calorific value of the power supply can be obtained by the following equation using the internal loss of the power supply and time.
・Heat generation (J) = Internal loss of power supply (W) x Time (s)
　　　　　　　　　  = [Effective power of input (W) – Output power (W)] x Time (s)
　　　　　　　　　  = {[Output voltage (V) x Output current (A) / Efficiency] - [Output voltage (V) x Output current (A)]} x Time (s)

Calculate the amount of heat generated per second based on the following operating conditions (example).
・Output voltage　24 V
・Output current　10 A
・Efficiency　　　 80%
・Time　　　　　 1 second

Heat generation (J) = {[Output voltage x Output current / Efficiency] - [Output voltage x Output current]} x Time (s)
　　　　　　　　  = {[24(V) x 10(A) / 80%] - [24(V) x 10(A)]} x 1(s)
　　　　　　　　  = [300(W) - 240(W)] x 1(s)
　　　　　　　　  =60 (W x s)
　　　　　　　　  =60 (J)
　　　　　　　　  =252 (cal)

For the power supply's efficiency characteristics, refer to the static characteristics "Efficiency vs. Output current" in the Evaluation Data of the technical data for each product.