# Tricky problem involving power supply efficiency

I have the following problem.

My attempt

Initally the efficiency of the supply is 83%, so $$\\dfrac{P_{out}}{P_{in}} =83\%\$$.

The remaining 17% must be power lost in dissipation in the enclosure $$\\dfrac{P_{diss}}{P_{in}} = 17\%\$$

If the enclosure is dissipating 25W then we have $$\P' = \dfrac{25\text{W}}{17\%} = 1.47 \frac{\text{W}}{\%} \$$

So the output power must be $$\P_{out} = 1.47\frac{\text{W}}{\%} \cdot 83\% = 122.01\text{W} \$$

If the efficiency is increased to 92% we have $$\\dfrac{P_{out}}{P_{in}} =92\% \:\$$ and $$\\dfrac{P_{diss}}{P_{in}} = 8\%\$$

If the enclosure is dissipating 25W as before we have $$\P' = \dfrac{25\text{W}}{8\%} = 3.125 \frac{\text{W}}{\%} \$$

And the output power becomes $$\P_{out} = 3.125\frac{\text{W}}{\%} \cdot 92\% = 287.5\text{W} \$$

So the percent change of the maximum output power must be $$\text{percent change} = \frac{P_{out,2}-P_{out,1}}{P_{out,1}} \cdot 100\% = 135.54\%$$

That was a lot of calculations (I apologize for that) and it leads me to believe that there is an easier way of solving this. Furthermore, am I really finding the maximum output power if I set $$\P_{diss}=25\text{W} \$$? Intuitively I would think the maximum output power occured if $$\P_{diss} = 0\text{W}\$$.

• Maybe it helps if you realize that 3.125 W/% could also be expressed as 312.5 W. That's the input power of the supply (you named it that way in your first equation even). 25 W will go as heat to the case. The rest is the output (easy subtraction). Feb 28, 2022 at 12:12

The power output in scenario 1 is $$\25\text{ watts}\times \frac{83}{100-83}\$$ = 122.06 watts.
The power output in scenario 2 is $$\25\text{ watts}\times \frac{92}{100-92}\$$ = 287.5 watts.
Intuitively I would think the maximum output power occurred if $$\P_{diss} = 0W\$$