Calculating Power Dissipation in a Simple Circuit

  • Thread starter Thread starter BuBbLeS01
  • Start date Start date
  • Tags Tags
    Power
AI Thread Summary
The discussion revolves around calculating power dissipation in a circuit with two resistors, R1 and R2, given a voltage of 6 V. The initial calculation attempted was incorrect, leading to confusion over the use of formulas for power and current. Participants clarify that power can be calculated using P = I^2R and emphasize the importance of understanding voltage drops across resistors. The concept of final and initial values in power calculations is also highlighted, with a focus on correctly applying Ohm's Law. Ultimately, the thread seeks to resolve misunderstandings about power dissipation and the correct approach to circuit analysis.
BuBbLeS01
Messages
602
Reaction score
0

Homework Statement


2wn8sar.jpg


How much power is dissipated by R1? (R1=8 Ω, R2=17 Ω, V=6 V.)

Sorry the picture is so big!

Homework Equations





The Attempt at a Solution



P = E^2/R = 8^2/6 = 10.67W what am I doing wrong?
 
Physics news on Phys.org
Do I have to add the resistors??
 
No P=IV so the change in power is the dissipation of power so delta P=IV-IV
so use v=IR and get P=I^2R -I^2R for the points after and at (before) the resistor, also remember always final-initial
 
So I would do...
I = E/Req = 6/25
P = 0.24^2 * 8 - 0.24^2 * 17 = -0.5184
Negative? Why would it be negative?
 
Can someone explain this...

No P=IV so the change in power is the dissipation of power so delta P=IV-IV
so use v=IR and get P=I^2R -I^2R for the points after and at (before) the resistor, also remember always final-initial

I don't understand what Vi and Vf are?
 
so i need to multiply the voltage drop across the resistors by the current. but how do i calculate the voltage drop?
 
solved...
 
Back
Top