Power dissipation in resistors

[charger9198]

Calculate the power dissipated in R1, R2 and R



[/b]

Previously i worked out the Current I1, I2 and I3 so i planned to use this to calculate the power dissipated;

For current i got;

I1= 0.8409 A
I2= -0.4545 A
I3= 0.39 A

I used I^2/R to get the following answers

R1 = 0.8409^2 / 1 = 0.7071 W
R2 = -0.4545^2 / 2 = -0.4131 W
R3 = 0.39^2 / 3 = 0.4563 W

 

[Jobrag]

Think about these I*V=watts V=I*R

 

[gneill]

R1 = 0.8409^2 / 1 = 0.7071 W
R2 = -0.4545^2 / 2 = -0.4131 W
R3 = 0.39^2 / 3 = 0.4563 W

A negative real number squared gives a positive result!

 

[charger9198]

Of course...

R1 = 0.8409^2 / 1 = 0.7071 W
R2 = -0.4545^2 / 2 = 0.2066 W
R3 = 0.39^2 / 3 = 0.4563 W

Am i on the right lines with this?

 

[Jobrag]

You were right first time I should have checked but your notation fooled me / usually means divide * is multiply.

 

[gneill]

First, the power is p = V²/R or p = I²R. Since you are using currents and resistances the second equation applies here. So you should be MULTIPLYING the square of the currents by the resistance values, not dividing as you've indicated. Recheck the results with this in mind.

 

[charger9198]

so

p=(I^2)R

thus

(0.8409^2)*1 = 0.70711 W
(-0.4545^2)*2 = 0.41314 W
(0.39^2)*3 = 0.4563 W

Hopefully that's it!

 

[gneill]

so

p=(I^2)R

thus

(0.8409^2)*1 = 0.70711 W
(-0.4545^2)*2 = 0.41314 W
(0.39^2)*3 = 0.4563 W

Hopefully that's it!

Yes, that looks good for the given resistances and currents.

 

[charger9198]

Thanks for your help

 

[stemurdo]

charger, can you show me how you worked out those currents?

 

[NascentOxygen]

charger, can you show me how you worked out those currents?

Discussion of this circuit in this thread.