Homework Help: Confused with circuit that has two opposing emf

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1. Aug 11, 2015

toforfiltum

1. The problem statement, all variables and given/known data

Calculate the total power generated by G
2. Relevant equations
P=VI

3. The attempt at a solution
I used the P=VI formula to calculate the total power generated by G which is 40W, and the answer turns out to be correct. But actually, I don't quite understand this topic of opposing emf. Like, why I can't I use the formula of P= V2/R instead? Why can't I get the same answer? For when using this equation, I know that I musn't include the value of 0.5Ω, which is the internal resistance of G. But why can't I get the same answer?

Same goes for P=I2R. These three formulas all give 3 different answers, though I thought they are supposed to give the same value of P. What is wrong with my understanding?

2. Aug 11, 2015

Hesch

PG = 40W which must be consumed by "somebody": Here they are:

- The battery consumes V*I = ( EB + RB*I ) * I = EB * I + RB * I2 ( you may call EB*I the charging power ).

- The internal resistor in G consumes RG * I2.

- Rightmost resistor, R, consumes R * I2.

Now, try to calculate the value of R and balance the account.

Last edited: Aug 11, 2015
3. Aug 11, 2015

Hesch

EB is included in the loop and you haven't calculated the value of R yet.

But as for the power consumed by the resistors you could say:

P = V2/R = ( EG - EB )2 / ( RG + RB + R )

Last edited: Aug 11, 2015
4. Aug 12, 2015

ehild

The power generated by a battery is its emf multiplied by the current flowing through it.
P=IV= V2/R = I2R is the power consumed by a resistor R, if I current flows through it and V=RI is the voltage across it.

5. Aug 12, 2015

toforfiltum

Ah, so I see that the P=VI, P=V2/R and P=I2R only applies to power dissipated by the loads in the circuit. Now I see where I'm wrong. Is it so because the V in the equations refer to the potential difference across circuit?

6. Aug 12, 2015

Hesch

Different components in the loop have different behaviour as to power consumption.

The battery has a constant emf ( 12V ), thus Pbat = V * I.

The resistors have a back-emf = R * I, thus

Pres = V * I = ( R * I ) * I = R * I2

or

Pres = V * I = V * ( V / R ) = V2 / R.

7. Aug 12, 2015

toforfiltum

Actually, I am not too clear about potential difference across terminals of batteries in these types of cases. According to what I have read, the terminal potential difference across generator G will be E- Ir, where E=20V and Ir= 2×0.5. This value is the same as say, when generator G is the only single source of emf in circuit. It doesn't seem too logical to me. For in another circuit where generator G is the single source of emf, the net emf is the emf of G itself, whereas in this circuit it is emf of G - that of battery.
How is this possible?

8. Aug 12, 2015

toforfiltum

What does back emf mean?

9. Aug 12, 2015

Hesch

It's the (counter) voltage across the resistor: Vres = R * I. ( Ohms law ).

10. Aug 12, 2015

BvU

@Tofo: Is it clear to you that the battery is being charged ? With a current of two amps electric power is "pumped in" to the tune of (12 V + 2 A * 0.1Ω) * 2 A (and being converted into chemical energy and heat). So the battery dissipates a big part of the electric energy G delivers; the remainder goes into the other two resistances.

: mistyped; sorry for this spurious alert, user Toto

11. Aug 12, 2015

toforfiltum

Haha, you still got my username wrongly. It's not Toto, though Tofo reminds me of Tofu instead.
Well. I think I'm clear that the battery is being charged. Anyway I have some confusions to clear up, and if you can help me see post #7.
Anyway, I'm curious to know what makes you think I'm unclear about the battery being charged?

12. Aug 12, 2015

toforfiltum

Why is there a counter voltage? Sorry, electric circuits is not my strongest topic. Is it because of the battery?

13. Aug 12, 2015

toforfiltum

@Hesch Ah I think I know what you mean now. The back emf is the voltage drop due to resistance of resistor, is it not?

14. Aug 12, 2015

BvU

There is a counter voltage because of the chemical (actually: physical ) processes in the battery that generate a potential difference at the terminals.
Oh sorry, that wasn't about the battery.

And yes, you're right about the 'counter' voltage over the resistors. They 'resist' the current that is being pushed through !

15. Aug 12, 2015

toforfiltum

Ahh, great! Turns out that 'counter' voltage is just resistance after all.... Been wondering if it's a new physics concept or something, sheesh.

16. Aug 12, 2015

BvU

Yes, over the terminals of "G&0.5Ω" a voltmeter will show 19 V.
But over the terminals of the battery plus the 0.1Ω a voltmeter will show 12.2 V in the given circuit.
And you already calculated the value of R to make it work as described ?

The term 'net emf' is a bit vague. Who said the net emf in this circuit is emf of G (20 V I suppose you mean) minus that of the battery (12 V ?) ? What is true is that the three resistors in series have to deal with a total potential difference of 8 V. And yes, there is an emf of 12 V in the circuit; a potential difference in the opposite direction of the current.

17. Aug 12, 2015

toforfiltum

Yes, the value of R is 3.4Ω.

Errm, I took it as an example from my textbook, though both circuits are not entirely the same. If I were to follow the example in my textbook closely, in the circuit avove there would be no R, just G and the battery. Does it make any difference?

The presence of this emf acts as a resistor and also to reduce the circuit's pd or emf (not sure) is it? Are these two things the only significance of the presence of this opposing emf?

18. Aug 12, 2015

toforfiltum

Wait, if 8V is the total potential difference in the circuit, why must we use the 20V value when calculating terminal potential difference across G? Why not 8V?

19. Aug 12, 2015

ehild

Imagine a resistor which is a simple straight wire of length L. The charge is driven across a resistor by the electric field E inside the resistor wire. While the charge moves from one end to the other, work is done by that field, and the work is equal to W=EqL. The work done on unit charge is equal to the potential difference across the wire. In practice, we call the negative of the potential difference "voltage" V. So the work by the electric field inside the resistor on charge q is W=qV. That work is lost, it transforms into heat, the resistor "dissipates" it. The power is the work done in unit time, and the charge flowing through the resistor in unit time is the current, I. So the power dissipated by the resistor is P=VI. According to Ohm's Law, V=RI. So the formula for the power dissipated by a resistor R can be used in three forms: P=VI=V2/R=I2R.

In a battery, there is a chemical reaction that separates the charges of opposite sign and cumulates them on the terminals. The chemical work is opposite to the electric work of the field between the opposite charges.
As the terminals of the battery have different charges, there is a potential difference across them. The potential difference across an unloaded battery is called emf - the electromotive force of the battery, denoted by ε .
When the battery is loaded, we get back the chemical energy as electric work done on the charge flowing through all the resistors in the circuit, including the internal resistance of the battery.
When current flows, the charge on the terminals would decrease, but the chemical reaction produces new charges continuously. The emf of the battery is kept constant, and the chemical work is W = ε * q, the power is P=ε*I, the power generated by the generator.

When you use the term emf it means electrical potential difference due to some non-electrical process. The voltage across a resistor is not "back emf ".

20. Aug 12, 2015

ehild

The total potential difference in a loop is zero. The 8 V is the net potential difference across all resistors. The problem asks the power generated - supplied - by the 20 V generator.

21. Aug 13, 2015

toforfiltum

What does that mean? How is that possible?
And thanks for giving such a detailed explanation.

22. Aug 13, 2015

ehild

23. Aug 13, 2015

toforfiltum

Thanks. I have another question. If there are two opposing emf sources in a circuit, and one source is greater than the other, will there be current from the smaller emf source?I don't think so. Am I right?
And if so, will the emf of the smaller source become like a resistor and cause a voltage drop that is equal to its emf value?

24. Aug 13, 2015

BvU

You 've got it right.

My advice: take a few AA batteries and a voltmeter (or a bicycle light bulb or something) and play with them.

25. Aug 13, 2015

William White

An easy way to do this is to use superposition.

Make sure you label the source correctly ( + or -)

Remove the first source, and solve for the current. You will get a positive or negative current depending on the "direction" of the source (ie what way round the battery is connected)

Then remove the first source, and add the second. Then solve for current The current will be flowing in the opposite direction.

Then add them together

So if your current for the first source was -15A
and for the second, was +10A

The total is -5A so the current is flowing from the first source to the second