# Total Dissipated Power

Please help look into to the below question where total dissipated power is to be calculated. In the solution given in the text they have added the under root of the power and then squared the result.

However, as per my understanding according to the superposition theorem you are not allowed to add the power together as it is not a linear quantity. Can someone help me explain why the powers are added the way they are in the solution is this an accepted formula. Thanks in advance.  However, as per my understanding according to the superposition theorem you are not allowed to add the power together as it is not a linear quantity.

They are not adding power together. They are adding sqrt(power). Sqrt(power) is linear under superposition.

Sqrt(power) has the dimensions of k.voltage or k.current, where k is a suitable bit of resistance.

• I've never come across that before, @Neil. Would k not be $\sqrt R$ (since you've un-squared the voltage or current)? – Transistor Sep 27 '19 at 9:48
• @Transistor one of them would. The other k would be 1/sqrt(R) – Neil_UK Sep 27 '19 at 9:49

Power is proportional to $$\V^2\$$ or to $$\I^2\$$, so the square root is taken in order to add these quantities by superposition. The result is then squared to return the answer to the power domain.

However, this question doesn't take into account that the power spec doesn't tell you the polarity of the current or voltage. This is accounted for mathematically by noting that the square root may be positive or negative. As such, a) 1 is also a potentially correct answer.

Not sure why they tried to include 5W twice in their answer list, unless they were trying to signal to the reader that this wasn't the right answer.