# What is the difference between the potential difference, and its emf?

JaeSun
ok, i guess i didnt see this forum ... i posted in another forum my question:

ok, im taking engineering physics II ... unfortunately, im forced to take this as part of my COMPUTER SCIENCE degree ... (why i have no idea..stupid, as i dont really care how many newtons of force it takes me to push a key on the keyboard, and how much current it takes to go from the keyboard to the computer to put the key on the screen) .... sorry, just im not the most science-oriented person. which is why im not a computer engineer.

so i have a question.

the question on my homework is this:

A sloar cell generates a potential difference of 0.10V when a 500-Ohm resistor is connected across it and a potential difference of 0.16V when a 1000-Ohm resistor is substituted. What are (a) the internal resistance and (b) the emf of the sloar cell? (c) The area of the cell is 5.0 (cm^2). What is the efficiency of the cell for converting light energy to internal energy in the 1000-Ohm external resistor?

question. What is the difference between the potential difference, and its emf? i know V=IR . isnt the total V what comes out of the power source? and isnt the emf the power source? I'm lost on what this is.

also, for part c ... how would i approach in solving this? The chapter deals with the circuits, resistors, etc. but effeciency?

any help would be appreciated!
thanks

## Answers and Replies

JaeSun
i know you guys dont just give answers...so im hoping you can point me in right direction?

here is what im thinking for (a) and (b) :

to find the internal resistance and emf, there are 2 variables, thus need 2 equations ....

so i set it up like this, according to the equations for internal resistance:

[del]V = &epsilon; * [ (R)/(R + r) ]

where R is the resistance of external circuit, and r is resistance of internal circuit. ***

***derived from V=IR , and for the internal resistance, I = &epsilon;/(R+r)

i guess since i have to find &epsilon; too, that i can use this equation.

i set-up two equations, according to whats given:

0.10V = &epsilon; [ (500)/(500+r) ]
0.16V = &epsilon; [ (1000)/(1000+r) ]

is that what i do? solve for one of the variables, input into the other?

i did that, and the math looks hard (hahaha).

using the first equation, and solving for &epsilon; , i get:

&epsilon; = .10 * [ (500+r)/500 ]

and inputting that into the other, makes it more confusing, to where i am stuck:

0.16 = [ ( 10 + r/50 ) ( 1 + 1000/r) ]
-or-
0.16 = 30 + 10,000/r + r/50

i am soooooooooooo lost. i have no idea if im going in the right direction or not ??

HELP!! this is due tomorrow!

gnome
We're just doing this now in my physics class too.

&epsilon; is the open circuit voltage. This is what the voltage drop across the power source would be when the current is zero (which is also what it would be if the internal resistance was zero).

That's where that formula you posted comes from -- in the form
&epsilon; = IR + Ir , where R is external resistance and r is internal resistance

But remember that &Delta;V, the net voltage drop across the source also equals the voltage drop across the external resistance so
&Delta;V = IR

So in your problem you can say
&Delta;V1 = .1V = I1R1
.1 = I1* 500&Omega;
I1 = .1/500 = .0002 A

and
&Delta;V2 = .16V = I2R2
.16 = I1* 1000&Omega;
I2 = .16/1000 = .00016 A

Now use the other equation
&epsilon; = IR + Ir
remembering that &epsilon; and r will be the same in both cases, so you can say
I1R1 + I1r = I2R2 + I2r
and solve for r.
And then you can solve for &epsilon;.

Then, as for the efficiency, I don't remember that being mentioned in class, and I don't see it in my textbook either. But I would guess that the efficiency is &Delta;V/&epsilon;. So, unless you come up with something more authoritative, I'd take a chance on that.

JaeSun
geez, never thought about solving the whole equation out and work from there....our book SUCKS, and it gave us that equation ... grrr...

i came out close though ... when i did it the long way, &epsilon;
came out as .2 , while doing it your way, was .4 ... it was close enough ...

thanks !!!!