Difference between power formulas

[mirandab17]

Hey there,

I'm very confused between the difference of these power formulas.

P= VI
P= (I^2)(R)
P=(V^2)/(R)

Why is it that I often get different answers with all of them? Don't they all equal power? What are the situations I can use each one in?

 

[gneill]

Hey there,

I'm very confused between the difference of these power formulas.

P= VI
P= (I^2)(R)
P=(V^2)/(R)

Why is it that I often get different answers with all of them? Don't they all equal power? What are the situations I can use each one in?

They should all give the same result if you are using the voltage V across a resistor R due the a current I passing through it.

 

[PeterO]

Hey there,

I'm very confused between the difference of these power formulas.

P= VI
P= (I^2)(R)
P=(V^2)/(R)

Why is it that I often get different answers with all of them? Don't they all equal power? What are the situations I can use each one in?

The most common reason for getting different answers is people using the incorrect value for V

Many people call V the Voltage [but never refer to weight as kilogramage or poundage, nor length as metreage or footage] and lose track of what it means

V is the Potential difference.

In MANY applications we do not know the value of V.

eg Power loss in the cord of a domestic electric kettle: here in Australia the supply Voltage is 240V [ie the Potential Difference across the socket is 240V]
When the kettle is turned on, there is a possibility that 0.1 V is dropped across the cord, and the other 239.9 across the element of the kettle. But that 0.1 was only a guess and could be completely wrong!
It is relatively easy to measure the resistance of the element in the kettle, and the current being drawn by the kettle. That is why we frequently use the P = I²R formula for power loss.
NOTE: Once we know the power loss, the other two formulae are useful if you want to go back and calculate what the Voltage Drop across the cord actually was.

 

[mirandab17]

Well the thing is, for example in this question: http://kss.sd69.bc.ca/pages/courses/Physics 12/Phys12Exams/aug02key.pdf
It's #6 written.

I did everything correctly but then, instead of calculating the current going through the 6 ohm resister and then using E=VIt to solve, I used V^2/R to find the power which is the same as P=VI. However, my answer was wrong when I did that.

All I did was changed the calculation for power which should be equivalent...

 

[gneill]

Well the thing is, for example in this question: http://kss.sd69.bc.ca/pages/courses/Physics 12/Phys12Exams/aug02key.pdf
It's #6 written.

I did everything correctly but then, instead of calculating the current going through the 6 ohm resister and then using E=VIt to solve, I used V^2/R to find the power which is the same as P=VI. However, my answer was wrong when I did that.

All I did was changed the calculation for power which should be equivalent...

Can you reproduce your calculations here? Then we can see if what you did was correct, or if not, where things went off the rails

 

[mirandab17]

Yeah sure!

Okay so first off
Rp = 1/(1/16 + 1/9) = 23.6
Then current of total circuit.
I = V/R = 12/23.6 = .5085 A

Then voltage of each series part.
V1 = IR = (.5085)(15) = 7.63 V
V2 = IR = (.5085)(5) = 2.54 V

Then to find voltage parallel.
Vp = 12 - 2.54 - 7.63 = 1.83 V

Then I thought I could go straight into E=Pt with my power calculation.

P = V^2/R = (1.83)^2/6 = .55663 W
E = (.55663)(15) = 8.4 J

...oh wow, I got it right this time... heheheh sorry. Although multiple times in the past I would get power calculations wrong. What exactly does V represent? Any ideas on why this would happen?

 

[mirandab17]

AKA this one:

http://kss.sd69.bc.ca/pages/courses/Physics 12/Phys12Exams/jan03.pdf
#22
Prime example. Don't know how to go about it at all.

 

[PeterO]

AKA this one:

http://kss.sd69.bc.ca/pages/courses/Physics 12/Phys12Exams/jan03.pdf
#22
Prime example. Don't know how to go about it at all.

The left branch obviously has resistance 16Ω, the right branch has a resistance of slightly more.

From that you can calculate how the 5A splits up [slightly more than 2.5A on the left and slightly less than 2.5 on the right.

That Right hand current splits into two equal parts for the passage through the 20Ω resistors.

Once you know that: use P = I²R

Alternately, use a few more minutes to calculate the Potential drop (V) across the 20Ω resistors and use P = VI or P = v²/R

 

[gneill]

Yeah sure!

Okay so first off
Rp = 1/(1/16 + 1/9) = 23.6 <----Careful, that 16 should be a 6, and the Rp should be 3.6 Ω.
Then current of total circuit.
I = V/R = 12/23.6 = .5085 A <---- yes!

Then voltage of each series part.
V1 = IR = (.5085)(15) = 7.63 V
V2 = IR = (.5085)(5) = 2.54 V

Then to find voltage parallel.
Vp = 12 - 2.54 - 7.63 = 1.83 V <---- Yup!

Then I thought I could go straight into E=Pt with my power calculation.

P = V^2/R = (1.83)^2/6 = .55663 W
E = (.55663)(15) = 8.4 J <---- Yessir!

...oh wow, I got it right this time... heheheh sorry. Although multiple times in the past I would get power calculations wrong. What exactly does V represent? Any ideas on why this would happen?

The V required is the potential difference across the resistance for which you wish to find the power dissipation. Ohm's Law connects the resistance, voltage, and current.

 

[mirandab17]

Thanks Gneil, you've been a lot of help!

As for #22, how would I go about calculating the current splitting up between the two branches? :S

 

[gneill]

Thanks Gneil, you've been a lot of help!

As for #22, how would I go about calculating the current splitting up between the two branches? :S

If I wanted a calculated value I would first note that the two parallel 20 Ω resistors make up 10 Ω together, so adding the single 10 resistor above them gives 20 Ω total for that branch. Then apply the current division rule for parallel branches:

$I_2 = 5.0A \frac{16}{16 + 20}$

(current splits according to the ratio of the resistance of the "other" branch to the sum of the resistances of the branches)

 

[PeterO]

Thanks Gneil, you've been a lot of help!

As for #22, how would I go about calculating the current splitting up between the two branches? :S

You could go one step further with the 16Ω and total of 20Ω in parallel and find the effective resistance, to see that the effective resistance of the lot is 9.88888..

For 5A to be flowing, that means the PD across the bunch is 49.44444.. Volts.

That means when the PD is split 50-50 in the right branch we have 24.722222.. across the 20Ω resistor - so use P = V²/R

 

[mirandab17]

Wow, definitely did not know the current division rule. Thanks for clearing that up!

GNeil, your explanation also makes a lot of sense too. http://kss.sd69.bc.ca/pages/courses/Physics 12/Phys12Exams/jun04.pdf

#22 is giving me a tough time here again if any of you could help me out.

Is the "voltage loss due to the battery’s internal resistance" equivalent to terminal voltage?
If so, I would do the equation like so but I keep getting the same answer.

First, I would find the current going through the system.
E(emf)= V(terminal) + Ir
20 = 2 + I(1.5)
I= 12

Then I was thinking I could use V = IR, the V being the EMF value and then the resistance being total resistance, including internal resistance.

V = IR
20 = (12)(1.5 + R)
R = .167 which is VERY far off.

...help please... :S

 

[gneill]

http://kss.sd69.bc.ca/pages/courses/Physics 12/Phys12Exams/jun04.pdf

#22 is giving me a tough time here again if any of you could help me out.

Is the "voltage loss due to the battery’s internal resistance" equivalent to terminal voltage?

No, it's the voltage that is dropped inside the battery due to its internal resistance -- it's voltage that's "lost" to the user. In this case you're told that this loss amounts to 2.0V which is dropped by the internal 1.5 Ω resistor. Looks like a job for Ohm's law...

If so, I would do the equation like so but I keep getting the same answer.

First, I would find the current going through the system.
E(emf)= V(terminal) + Ir
20 = 2 + I(1.5)
I= 12

Then I was thinking I could use V = IR, the V being the EMF value and then the resistance being total resistance, including internal resistance.

V = IR
20 = (12)(1.5 + R)
R = .167 which is VERY far off.

...help please... :S

 

[mirandab17]

Gah, sorry, I still don't get it. What variable does the 2.0 V represent then?
Can I find the current of the system by
E = IR
20 = I (1.5)
I= 13.3 A?

 

[gneill]

Gah, sorry, I still don't get it. What variable does the 2.0 V represent then?
Can I find the current of the system by
E = IR
20 = I (1.5)
I= 13.3 A?

Nope. The 2V is dropped across the 1.5 Ω resistance. So what current must flow through 1.5 Ω in order to drop 2V?

 

[mirandab17]

I'm assuming I would do I = V/R so I = 1.33.
I always thought that I would use the emf value for that. Interesting... So is this the total current of the circuit then?
Couldn't I then do E = IR so 20 = 1.3 (R + 1.5).

Oooh that's the right answer now!

 

[gneill]

I'm assuming I would do I = V/R so I = 1.33.
I always thought that I would use the emf value for that. Interesting... So is this the total current of the circuit then?
Couldn't I then do E = IR so 20 = 1.3 (R + 1.5).

Oooh that's the right answer now!