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Power dissipated by a resistor in combined citcuit

  • #1
6
0
Find the power dissipated by R2 given that V a-b=120 volts, R1=1, R2=3.33, R3=2, and R4=3

I've worked this problem, just want to confirm I worked it correctly.

A---------R1-----------*-------------------
| |
R2 R3
| |
B---------R4----------*--------------------

Req=5.25 Ohms, therefore total current, I, =V/Req 120V/2.25 Ohms I=22.86A
We then find the voltage drop over R1, since it sees all the current in the circuit
V=IR V=22.86A*1 Ohm v=22.86
we also know that R4 will see all current in the circuit, so we calculate its voltage drop as well.
v=68.58
the total of these two drops subtracted from the total equals 28.56.
Since R2 and R3 are in parallel, they see equal current, meaning the remaining voltage is dissipated over these two resistors.

using the power formula, P=V^2/R, we can find the power dissipated by R2 by 28.56^2/3.33 P=244.95 watts.

Is this correct, or have I missed something somewhere
 

Answers and Replies

  • #2
6
0
Just a note, the R2 is inbetween the two stars and R3 is at the end of the circuit. SOrry for the confusion
 
  • #3
gneill
Mentor
20,792
2,770
Looks fine. Note that you should avoid rounding intermediate results except for presentation purposes; keep extra digits in intermediate values that will be used in further calculations to keep rounding errors from creeping into significant digits in final results.

If you want to post ascii diagrams, consider placing their text between [ code ] [ /code ] tags. That way spacing and indents will be preserved.
 
  • #4
CWatters
Science Advisor
Homework Helper
Gold Member
10,529
2,295
Looks ok but..

Since R2 and R3 are in parallel, they see equal current voltage, meaning the remaining voltage is dissipated over these two resistors.
 

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