Comparing EMF and Voltage in Circuit Systems

[Acuben]

First of all this isn't really Homework problem, but since it is based on homework problem, I'll post on this section of the forum. Feel free to move if I was supposed to post on different part of the forum.
Either way, I'll use homework as an example since that is what brought this up..

Regardless of how long this post looks, it's really a simple question =p

Homework Statement

This question involves circuit system using...
Power, Voltage, Current, and Resistance (internal and load resistance), and EMF
This is probably really basic question for most of you

Question is when measuring Current and Power of any part of the circuit, do I use EMF? or Voltage? (So it's two questions. 1. when measuring current, 2. when measuring power)
say...
P=V^2 / R or P=E^2 / R ?
and I= V / R or I = E / R ?
and when using other Power formula such as P=I*V, P=I^2 * R
do I use the current that is derived from using EMF ? or Voltage?
assuming I use I = E/ Rt

Homework Equations

P=I*V=V^2/R=I^2*R
I=V/R
V=E-Ir (or V=E+Ir)

where...
P-power
I-current
V-voltage
R-Resistance
Rl-Load Resistance
Ri- Internal resistance
Rt-total resistance
E-EMF (Electro Motive Force ithink...)

The Attempt at a Solution

This is a conceptual question

Don't read this part, unless you feel it's neccessary...It's not really quotation, I just wanted to box it somehow ...

it seems like there is a controversy in the textbook or I simply have wrong idea about this Voltage vs EMF deal...
Let's not worry about significant figures and units in detail...

Book Physics for scientists and Engineers with Modern Physics by Serway Jewett Volume 2, 6th ed (older edition afaik)
page 861
one of the example they gave was...
E= 12.0V
Ri=0.05
Rl=3.00
I= E / (Ri + Rl) = 120 / 3.05 = 3.93
(not I=V / (iR+Rl) ...)

page885 (homework section, number1)
battery with E= 15.0V
V(terminal)= 11.6V delivering 20.0W of power to Rl
a) what is value of Rl?
Rl=V^2 / P
so I have two choices...
1. If V= Voltage
Rl= 11.6^2 / 20
2. If V= EMF
Rl= E^2 / P
Rl= 15^2 / 20

so I was thinking. Of course it should be of course the answer should be E^2/P therefore 15^2/20 > 11.25 Amps
but the answer shows as 6.73 Ohms... = 11.6^2 / 20

so in example it used EMF
and in Homework it used terminal voltage.

So one of them should be wrong?
or is there something that I am not understanding?

Oh by the way, my professor did mention that he'll purposely assign homework with wrong answers just because :P

Is there something that I don't know and should know? (maybe about V vs V(terminal)?)

 

[lewando]

In 2010, I think the term "electromotive force" is at risk for becoming an archaic term. Sort of like "kilocycles" or "kc" for kiloHertz. I have led a happy life without being able to distinguish an EMF from a Volt.

Think of EMF as the "concept" equivalent to voltage measured in Volts.

 

[Acuben]

In 2010, I think the term "electromotive force" is at risk for becoming an archaic term. Sort of like "kilocycles" or "kc" for kiloHertz. I have led a happy life without being able to distinguish an EMF from a Volt.

Think of EMF as the "concept" equivalent to voltage measured in Volts.

I believe i am aware of what EMF is. correct me if I am wrong.
EMF is a type of V that's conceptually pure volt from (say a battery) unaffected by any resistance
whereas V(terminal) is type of V that is actually used. (more calculated)

it's not understanding EMF/ V(terminal) that I'm having trouble with. I'm having trouble with actually applying them in an equation.

 

[lewando]

EMF is voltage. You would do well to forget you ever heard of EMF. If you see the term "EMF" in a book, white-it-out and replace with ink the word "voltage". I'm not kidding.

Just use these formulas and you will be fine for all things ohmic:

V=IR
P=V²/R
P=I²R

 

[lewando]

page885 (homework section, number1)
battery with E= 15.0V
V(terminal)= 11.6V delivering 20.0W of power to Rl
a) what is value of Rl?
Rl=V^2 / P
so I have two choices...
1. If V= Voltage
Rl= 11.6^2 / 20
2. If V= EMF
Rl= E^2 / P
Rl= 15^2 / 20

so I was thinking. Of course it should be of course the answer should be E^2/P therefore 15^2/20 > 11.25 Amps
but the answer shows as 6.73 Ohms... = 11.6^2 / 20

I should spend more time actually reading original posts. Sorry. In this case you have a battery modeled as an ideal voltage source of 15V in series with an internal resistance.
In this case, they are simply telling you that the teminal voltage, Vt, is what is appearing across the load resistor and so P= Vt^2/R applies.

 

[Acuben]

@lewando
thank you for reply
so I use V(terminal) for calculating power and EMF for calculating total current?

 

[lewando]

For power dissipated by the load resistor, from the point of view of the load resistor-- Vt is applied across it so that's what you use. For current going through the resistor, why not use I=V/R with V = Vt and R = Rload?

Vemf can be used if you want to calculate the total power dissipation of the circuit (power dissipated by Rload and Rint), because Vemf is the voltage across the series combination of the two resistors)

 

[Acuben]

For power dissipated by the load resistor, from the point of view of the load resistor-- Vt is applied across it so that's what you use. For current going through the resistor, why not use I=V/R with V = Vt and R = Rload?

Vemf can be used if you want to calculate the total power dissipation of the circuit (power dissipated by Rload and Rint), because Vemf is the voltage across the series combination of the two resistors)

Ah I see it now. Thank you very much