Power Dissipated by Resistors

Discussion in 'Homework Help' started by Elizabeth Paredez, Mar 22, 2011.

1. Elizabeth Paredez Thread Starter New Member

Mar 22, 2011
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The only thing I know about resistors is that V = IR. I think I can simplify the circuit but I really have no idea how to answer the following questions.

What is the power dissipated by each of the resistors in the following circuit and what is the electric potential difference between points (a) and (b)?

Please any help would be greatly appreciated because I don't even know where to start.

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2. Wendy Moderator

Mar 24, 2008
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3,049
Take the two parallel resistors, make them one, then you will have two series resistors. Make them one, then you have it.

3. Elizabeth Paredez Thread Starter New Member

Mar 22, 2011
3
0
Here's what I did.

V = IR

I combined the 2.0 Ω resistor with the 24.0 Ω resistor to give me "26.0 Ω" resistor. Since I made it a parallel circuit, volts are constant.

For R(equivalent) I got (26^-1) + (8^-1) = .163461....then ^-1 it equals approx 6.0

Then I did, I26 = V26/ R26 (12/26) and got approximately .5 Amps
Then I did, I8 = V8/R8 (12/8) and got approximately 1.5 Amps

Then I did, I(total) = V/Req (12/6) and got approximately 2.0 Amps

Does anybody know if what I did is correct???

4. Wendy Moderator

Mar 24, 2008
21,839
3,049
You can't do it that way. You must combine the parallel, since both of them are in series with the 2Ω resistors.

Combine the two parallel resistors first.

5. Georacer Moderator

Nov 25, 2009
5,154
1,281
The reason you can't combine the 24Ω and the 2Ω resistor is that they are neither parallel nor series.
They would be parallel if they had common endpoints.
They would be series if they had only one common endpoint, which would be unique to the two of them.
In your case, that endpoint is shared with the 8Ω resistor which spoils the "serial-ity".

6. saikat36 Member

Jan 28, 2009
16
0
U have 2 parallel resistor 24? & 8?.So

Req=(R1*R2)/(R1+R2)=6?

Now 6? & 2? r in series, so
Req=6+2=8?
then current I=(12/8)=1.5A
now branch current of 24? & 8?

I24=(8/32)*1.5=.375A

I8=(24/32)*1.5=1.125A

Power dissipation

P24=24*0.375*0.375=3.375W
P8 =8*1.125*1.125 =10.125W
P2 =2*1.5*1.5 =4.5W

Vab=-(2*1.5)=-3V

I think now all is cleared.
Thank u.