Confused with circuit that has two opposing emf

[toforfiltum]

Homework Statement

Calculate the total power generated by G

Homework Equations

P=VI

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?

 

[Hesch]

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

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

- The internal resistor in G consumes R_G * I².

- Rightmost resistor, R, consumes R * I².

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

 

[Hesch]

why I can't I use the formula of P= V²/R instead?

E_B 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 = ( E_G - E_B )² / ( R_G + R_B + R )

 

[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.

 

[toforfiltum]

E_B 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 = ( E_G - E_B )² / ( R_G + R_B + R )

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.

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?

 

[Hesch]

Is it so because the V in the equations refer to the potential difference across circuit?

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

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

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

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

or

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

 

[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?

 

[toforfiltum]

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

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

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

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

or

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

What does back emf mean?

 

[Hesch]

What does back emf mean?

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

 

[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.

[edit]: mistyped; sorry for this spurious alert, user Toto

 

[toforfiltum]

@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.

[edit]: mistyped; sorry for this spurious alert, user Toto

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?

 

[toforfiltum]

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

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

 

[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?

 

[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 !

 

[toforfiltum]

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 !

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

 

[BvU]

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?

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.

 

[toforfiltum]

And you already calculated the value of R to make it work as described ?

Yes, the value of R is 3.4Ω.

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 ?) ?

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?

And yes, there is an emf of 12 V in the circuit; a potential difference in the opposite direction of the current.

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?

 

[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?

 

[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 ".

 

[ehild]

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?

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.

 

[toforfiltum]

The total potential difference in a loop is zero.

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

 

[ehild]

Recall Kirchhoff's Voltage Law (KVL) . http://www.electrical4u.com/kirchhoff-current-law-and-kirchhoff-voltage-law/

 

[toforfiltum]

Recall Kirchhoff's Voltage Law (KVL) . http://www.electrical4u.com/kirchhoff-current-law-and-kirchhoff-voltage-law/

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?

 

[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.

 

[William White]

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?

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

 

[toforfiltum]

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

Thanks, so to confirm in the circuit there is current flowing only in one direction right? The value on ammeter recorded is not the net current, right? Or not?

 

[toforfiltum]

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.

Thanks, but just after I posted this question, I saw in my textbook that the potential difference across a non passive resistor is equal to the sum of its emf and the voltage drop across its internal resistance. So...?

 

[William White]

The ammeter records the TOTAL current. And it is + or - depending upon which way you connect it

Consider the circuit attached (top diagram) I want to know what the total current is, and what way it is flowing.

It does not matter which way you chose is positive or negative right now.

So I remove the second source from the circuit (middle diagram). Assume the current is clockwise from the positive terminal to the resistor. The current is just 100V / 50 ohms = 2A

Then I do the same for the bottom circuit. Assumng the current is clockwise the current is just -200/50 = -4A (I have kept the convention that the current is going the same way; but this time the voltage is negative because the battery is connected 'backwards')

Add these together and I get -2A TOTAL

The minus means I might have assumed incorrectly the direction in the first instance So the current in the top circuit (the entire circuit) is +2A counter clockwise.If you have more than one source, this is often the easiest way to solve the circuit. It might take a little more time (because you have to solve the circuit for each source); but it makes things a lot simpler.

 

[William White]

Thanks, so to confirm in the circuit there is current flowing only in one direction right?

You are interested in the total current.

It has a single size and a single direction.

In my example the size is 2A and the direction is counter-clockwiseDon't forget, if you get a negative current, it is the same answer, you just drew your arrow the wrong way in the first place (like I did!). This is the same as in reality if you connected your ammeter backwards.

 

[toforfiltum]

The ammeter records the TOTAL current. And it is + or - depending upon which way you connect it

Consider the circuit attached (top diagram) I want to know what the total current is, and what way it is flowing.

It does not matter which way you chose is positive or negative right now.

So I remove the second source from the circuit (middle diagram). Assume the current is clockwise from the positive terminal to the resistor. The current is just 100V / 50 ohms = 2A

Then I do the same for the bottom circuit. Assumng the current is clockwise the current is just -200/50 = -4A (I have kept the convention that the current is going the same way; but this time the voltage is negative because the battery is connected 'backwards')

Add these together and I get -2A TOTAL

The minus means I might have assumed incorrectly the direction in the first instance So the current in the top circuit (the entire circuit) is +2A counter clockwise.If you have more than one source, this is often the easiest way to solve the circuit. It might take a little more time (because you have to solve the circuit for each source); but it makes things a lot simpler.

Hmm...then if to follow your logic, say I now have a circuit where the two opposing emfs are equal. Then, there would be a positive value of current and a negative value of current flowing in the circuit. The ammeter reading is zero. But if I were to put my fingers across the wire, I would feel a shock due o the current, no matter where it is flowing.
But it doesn't make much sense to me for current to flow in that circuit since there is no net potential difference across that loop.

 

[toforfiltum]

@William White Ah ok so current flows in one direction only. Ok thanks. My teacher had me thinking that there were two opposing flow of current in that circuit.

 

[William White]

@William White Ah ok so current flows in one direction only. Ok thanks. My teacher had me thinking that there were two opposing flow of current in that circuit.

I reality there is a sum of currents.

Imagine pushing a block forwards, and your friend pushing it backwards, but with a greater force than you. The total effect is that it will go backwards.

But to know that, you need to keep a track of both forwards and backwards forces.

So, you need to know the currents in both directions so you can know the total current and what direction it flows.

 

[toforfiltum]

I reality there is a sum of currents.

Imagine pushing a block forwards, and your friend pushing it backwards, but with a greater force than you. The total effect is that it will go backwards.

But to know that, you need to keep a track of both forwards and backwards forces.

So, you need to know the currents in both directions so you can know the total current and what direction it flows.

Ah I see. Ok thanks.

 

[William White]

Hmm...then if to follow your logic, say I now have a circuit where the two opposing emfs are equal. Then, there would be a positive value of current and a negative value of current flowing in the circuit. The ammeter reading is zero. But if I were to put my fingers across the wire, I would feel a shock due o the current, no matter where it is flowing.

If you touched a circuit where the sum of voltages was zero you would not feel a shock (as long as you were standing at zero voltage - say on the earth)
Seems counter-intuitive, but when you do AC theory you will see a more practical example. In 3-phase AC system, there are 3 voltages of equal magnitude, but they are 120° out of phase with each other (rather than 180° in your example with 2 sources). When you add them up, the total is zero. This is the neutral point. The currents flow from one line through the neutral into the other lines. As long as the voltages are equal and opposing it is quite safe to touch the neutral point. And in fact, this point is connected to Earth for safety.

 

[toforfiltum]

If you touched a circuit where the sum of voltages was zero you would not feel a shock (as long as you were standing at zero voltage - say on the earth)
Seems counter-intuitive, but when you do AC theory you will see a more practical example. In 3-phase AC system, there are 3 voltages of equal magnitude, but they are 120° out of phase with each other (rather than 180° in your example with 2 sources). When you add them up, the total is zero. This is the neutral point. As long as the voltages are equal and opposing it is quite safe to touch the neutral point. And in fact, this point is connected to Earth for safety.

Haha, this is out of my depth.I don't get the out of phase thing Guess I'm not there yet. Physics is hard to understand intuitively..