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Power dissipated?

  • #1

Homework Statement



What is the power dissipated by the 5.0Ω resistor in the following circuit?
4000123.gif

Homework Equations




I thought maybe to simplify the resistance like.

1/r=1/30 +1/30
r=15
+10+5
R=30ohms
then find the current
v=ir
25=Ix30
I=.8333
then find power with the 5 ohms?

P=vi
P=25x5
P=75w?
answer given is 6.2w?

thanks.
 

Answers and Replies

  • #2
161
0
Your current calculations look ok.

You have used the wrong value for voltage.

Remember that only a proportion of the 25V is used in the 5 ohm resistor.

I'll leave it up to you to figure out what the value should be.

Oh, and you've substituded wrongly also....
 
Last edited:
  • #3
diazona
Homework Helper
2,175
6
You can also find a formula for power in terms of current and resistance, instead of current and voltage as you have it.
 
  • #4
Im trying to think of what the voltage is at that point,
do I use,
EMF=Vab+IR ?
24= vab+0.83*5 or something?

I thought voltage was constant everywhere in a circuit?
 
Last edited:
  • #5
12
0
The voltage drop across loads connected in parallel is constant(equal to supplied voltage) whereas the voltage drop across loads connected in series is proportional to their resistance (V=IR)

So, calculate the voltage drop across the 5Ohm resistor and use that in you're calculation for Power.

Edit: I got 3.47W as the power dissipation though... I might have missed something.
 
  • #6
The voltage drop across loads connected in parallel is constant(equal to supplied voltage) whereas the voltage drop across loads connected in series is proportional to their resistance (V=IR)

So, calculate the voltage drop across the 5Ohm resistor and use that in you're calculation for Power.

Edit: I got 3.47W as the power dissipation though... I might have missed something.
your right its 3.47, the question was split over 2 pages so I saw the wrong answer.

I'm still kind of confused about the voltage when in parallel though.
so if you have a 20v power source that splits 4 ways each is 20v then as far as resistance you treat each on seperatly as if it was a series?
 
  • #7
Thanks.

so v=IR
25=Ix30
I=.833333

V=IR
V=.83x5
V=4.15

P=IV
P=4.15x.8333333
P=3.46


Is that right?
 
  • #8
12
0
Yes, that's how I did it.
 

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