Power Dissipation in Circuit Calculation

[brentwoodbc]

Homework Statement

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

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.

 

[tyco05]

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...

 

[diazona]

You can also find a formula for power in terms of current and resistance, instead of current and voltage as you have it.

 

[brentwoodbc]

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?

 

[Jiachao]

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.

 

[brentwoodbc]

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?

 

[brentwoodbc]

Thanks.

so v=IR
25=Ix30
I=.833333

V=IR
V=.83x5
V=4.15

P=IV
P=4.15x.8333333
P=3.46Is that right?

 

[Jiachao]

Yes, that's how I did it.