Power Calc

Okay, here's anohter dumb quesiton.

Say I have a resistor across the mains power supply. Say its 10k for
argeuments sake. Mains here is about 115VAC. So the resistor will
disspate 1.3 watts give or take by my rekoning. Now say I put a diode
in series with the resistor. Does the power dispataion now halve?

just curious.

Dan Green <dhg99908@hotmail.se> wrote:
> Okay, here's anohter dumb quesiton.
> 
> Say I have a resistor across the mains power supply. Say its 10k for
> argeuments sake. Mains here is about 115VAC. So the resistor will
> disspate 1.3 watts give or take by my rekoning. Now say I put a diode
> in series with the resistor. Does the power dispataion now halve?
> 
> just curious.
> 

Yes. The dissipation is the same in the positive and negative half-cycles,
so blocking the current in one direction halves the average dissipation. 

The diode drops some voltage when conducting, so the dissipation is
slightly less than half in real life. 

Cheers 

Phil Hobbs 

-- 
Dr Philip C D Hobbs  Principal Consultant  
ElectroOptical Innovations LLC / Hobbs ElectroOptics  Optics,
Electro-optics, Photonics, Analog Electronics

Dan Green wrote:
-----------------------------
> Okay, here's anohter dumb quesiton. 
> 
> Say I have a resistor across the mains power supply. Say its 10k for 
> argeuments sake. Mains here is about 115VAC. So the resistor will 
> disspate 1.3 watts give or take by my rekoning. Now say I put a diode 
> in series with the resistor. Does the power dispataion now halve? 
> 

**  Conducting for half the time, or any other fraction, reduces the heat in a resistor by that same fraction.
The  *average value* of the current through or voltage across the resistor is also halved but this is not so for the *RMS* value. RMS values follow the heating effect on a resistance so will be reduced by 0.7071  ( 1/sq rt 2)  or in your example from 115 down to 80.6 volts.



....  Phil