A circuit problem....Watch

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Thread starter 5 years ago
#1
I have been fortunate enough to borrow a load of circuit equipment and have experimenting at home. I have done a circuit that has really got me perplexed...

A 4 V cell leads to a 20 ohm resistor.

Thus:

V= 4V
R= 20 ohm
I = 0.2 A
Calculated resistor power rating (P=VI) = 0.8 W (so can dissipate 0.8 J/s)
The voltage across the resistor = 4 V (as expected from power rating)

I then proceeded to add in a 2.5 V lamp after the resistor in series

Thus:

V = 4V
I = 0.13 A
Total resistance = 30 ohms (approx)
Resistance of lamp = 10 ohms (approx)

The lamp lights up.... how does this happen? Surely with the power rating of the resistor there should be no surplus potential energy to light the lamp? I understand the lamp itself functions as a resistor and therefore slows the current but I just cant get my head round why the lamp lights up?!

Apols if Im missing something obvious!
0
5 years ago
#2
(Original post by Science hopeful)
I have been fortunate enough to borrow a load of circuit equipment and have experimenting at home. I have done a circuit that has really got me perplexed...

A 4 V cell leads to a 20 ohm resistor.

Thus:

V= 4V
R= 20 ohm
I = 0.2 A
Calculated resistor power rating (P=VI) = 0.8 W (so can dissipate 0.8 J/s)
The voltage across the resistor = 4 V (as expected from power rating)

I then proceeded to add in a 2.5 V lamp after the resistor in series

Thus:

V = 4V
I = 0.13 A
Total resistance = 30 ohms (approx)
Resistance of lamp = 10 ohms (approx)

The lamp lights up.... how does this happen? Surely with the power rating of the resistor there should be no surplus potential energy to light the lamp? I understand the lamp itself functions as a resistor and therefore slows the current but I just cant get my head round why the lamp lights up?!

Apols if Im missing something obvious!
Yes, you are missing the fact that you asked the same question here
http://www.thestudentroom.co.uk/show....php?t=2551022
and got the answer to that and this question.

If the lamp has 2.5V the resistor will have 1.5V to make up the 4V in series.
In general, the pd across the lamp and resistor will add to 4V. The pds will actually be in the ratio of the resistances of the lamp and resistor.
If the resistor is 20 Ohm and the lamp 10ohm the pds will be in the ratio 2 to 1.
0
5 years ago
#3
(Original post by Science hopeful)
calculated resistor power rating (P=VI) = 0.8 W (so can dissipate 0.8 J/s)
Power rating is NOT the same as power dissipated.

You calculated the actual power dissipated in the resistor from a measurement of the current and voltage:

P = VI
= 4 x 0.2
= 0.8W.

Power rating is the maximum power the resistor is able to safely dissipate before it is permanently damaged by excess heat.

(Original post by Science hopeful)
The voltage across the resistor = 4 V (as expected from power rating)
This is an incorrect statement. The voltage across the resistor is 4V because it is placed directly across the 4V battery. From ohms law:

The power dissipated in the resistor is a function of the voltage across the resistance. The voltage is the pressure which forces electrons through the resistance, i.e. causes a current to flow. The current in turn gives up energy and causes power to be dissipated given by:

I = V/R, therefore
I = 4/20
= 0.2A

P = VI
= 4x0.2
= 0.8W.

(Original post by Science hopeful)
Surely with the power rating of the resistor there should be no surplus potential energy to light the lamp?
This is incorrect thinking.

As stated previously, power rating is not the same as power dissipated.

With the lamp in series, you measured a new current of 0.13A. This new current flows through both the resistor and the lamp. i.e. the current is lower because the total series resistance has increased.

Rtotal = V/I = 4/0.13 = 31 ohms.

Vresistor = IRresistor = 0.13 x 20 = 2.6V

Vlamp = IRlamp = 0.13 x 11 = 1.4V

A stated previously, the power dissipated in the resistor is also reduced:

Presistor = Vresistor x Itotal = 2.6 x 0.13 = 0.34W

Plamp = Vlamp x Itotal = 1.4 x 0.13 = 0.18W

Notice that the total power dissipated in both the am and resistor is now also reduced.

Ptotal = Plamp + Presistor = 0.34 + 0.18 = 0.52W

(Original post by Science hopeful)
I understand the lamp itself functions as a resistor and therefore slows the current but I just cant get my head round why the lamp lights up?!
The current does not slow. Current is a measure of the quantity of electrons flowing in and out of the resistor in 1 second. Current does not depend on the speed of electrons.

The lamp lights because the energy carried by the electrons is now shared between the lamp and the resistor in the same ratio as their resistances.
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Thread starter 5 years ago
#4
(Original post by uberteknik)
Power rating is NOT the same as power dissipated.

You calculated the actual power dissipated in the resistor from a measurement of the current and voltage:

P = VI
= 4 x 0.2
= 0.8W.

Power rating is the maximum power the resistor is able to safely dissipate before it is permanently damaged by excess heat.
Sorry yes the resistor has dissipated 0.8W but that doesnt mean thats what its power rating was (ie chances are it is less than that)?

Power rating is a maximum figure

(Original post by uberteknik)
This is an incorrect statement. The voltage across the resistor is 4V because it is placed directly across the 4V battery. From ohms law:

The power dissipated in the resistor is a function of the voltage across the resistance. The voltage is the pressure which forces electrons through the resistance, i.e. causes a current to flow. The current in turn gives up energy and causes power to be dissipated given by:

I = V/R, therefore
I = 4/20
= 0.2A

P = VI
= 4x0.2
= 0.8W.
Understood

(Original post by uberteknik)

With the lamp in series, you measured a new current of 0.13A. This new current flows through both the resistor and the lamp. i.e. the current is lower because the total series resistance has increased.

Rtotal = V/I = 4/0.13 = 31 ohms.

Vresistor = IRresistor = 0.13 x 20 = 2.6V

Vlamp = IRlamp = 0.13 x 11 = 1.4V

A stated previously, the power dissipated in the resistor is also reduced:

Presistor = Vresistor x Itotal = 2.6 x 0.13 = 0.34W

Plamp = Vlamp x Itotal = 1.4 x 0.13 = 0.18W

Notice that the total power dissipated in both the am and resistor is now also reduced.

Ptotal = Plamp + Presistor = 0.34 + 0.18 = 0.52W
(Original post by uberteknik)
The current does not slow. Current is a measure of the quantity of electrons flowing in and out of the resistor in 1 second. Current does not depend on the speed of electrons.
OK.... I think this is where my problem is.

Excuse me for the personification here!....... How do the electrons 'know' to share their energy between the 2 resistors?
As demonstrated in the first circuit, the entire 4 J C-1 is dissipated by the resisitor, leaving zero voltage upon exiting. Why in the second circuit with the lamp is all the voltage not expended again in the resistor?

I understand why the current has reduced from 0.2 A to 0.13 A. Current is the number of coulombs that pass a point every second. So there are now fewer coulombs going through the resistor..... BUT each coulomb still has 4 J of energy? So why is it not expended as before.... how do the electrons 'know'?

The speed that each coulomb passes through the circuit does not change merely the quantity..... UNDERSTOOD!

In my mind I have an analogy of a current motorway.... one with 1 lane and the other with 5 lanes... the electrons on both motorways are traveling at light speed, its just the 5 lane motorway has more of them traveling at that speed?
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5 years ago
#5
(Original post by Science hopeful)
.....what its power rating was (ie chances are it is less than that)?
Designers will choose an appropriate resistor power rating form the manufacturers specifications so that the expected power dissipaton can be safely handled without overheating it. Typical power ratings are 0.25W, 0.5W, 1W, 2W, 5W etc.

(Original post by Science hopeful)
....... How do the electrons 'know' to share their energy between the 2 resistors?
OK. Apologies for the delay in responding, gotta have a life too! lol.

Voltage is always measured from a reference point. In the case of a power supply, that reference is the -ve battery terminal. This is labelled 0V or ground. The positive battery terminal will be at some voltage potential above that 0V reference. The +ve voltage potential defines the amount of joules per coulomb of charge the battery is able to supply to the circuit.

The keyword here is 'potential'. An analogy would be gravitational potential energy where an object placed on a ledge above the ground has potential energy. That g.p.e. will convert to kinetic energy if the object falls off the ledge. When it reaches the ground it has no potential energy left - it was all converted to kinetic energy.

If the falling object encounters a resistance on the way down, it may slow down but it carries on falling. More resistance and it falls even slower, but it will still convert all of it's g.p.e. until it reahes the bottom where the g.p.e. is zero. No matter what the resistance on the way down is, the object always had the same g.p.e. when it started falling and it will always be zero when it reaches the bottom and stops falling.

In a similar way, the voltage potential developed across any resistor, is a measure of the potential energy given up to that resistance. The voltage potential energy at the +ve terminal is a fixed value and as the electrons move through the circuit and meet resistance, they give up that potential energy at a rate governed by the resistance. i.e. power = joules /second converting it to heat in the resistor.

Adding more resistance in the path of the electrons and the rate of energy exchange is slowed. i.e. the current is reduced.

P = VI and since

V = IR (ohms law), substituting V in the power expression above gives

P = IR.I = I2R

hence the power dissipated (energy conversion rate) is governed by both current and resistance.

In essence the energy is exchanged uniformly throughout all of the resistance becuase the current must still flows through all of the resistance in it's path to get to the -ve terminal of the battery no matter what that value of resistance in between is.

Has this made sense?
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Thread starter 5 years ago
#6
Thanks for the reply.

I think I get it.... so using my first circuit as an example where there was just a resistor and a 4 V cell... When the voltage is measured across it, the actual figure isn't 4 V but ever so slightly less than that? The electrons still needing enough energy to get through the remaining wire (an example of an albeit extremely low resistor)? The resistor slows the current down significantly but it doesnt make it stop (like your falling object example?)

A greater number of coulombs passing through a point per second (ie) amperes going through the resistor) results in a greater amount of energy being transferred to the resistor and dissipated as heat due to the greater number of collisions?
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5 years ago
#7
(Original post by Science hopeful)
......When the voltage is measured across it, the actual figure isn't 4 V but ever so slightly less than that? The electrons still needing enough energy to get through the remaining wire (an example of an albeit extremely low resistor)?
You are overthinking and overcomplicating things.

I am just about to eat some dinner but will be on-line around 8.30pm GMT.

If you are on-line at that time, I'm sure we can crack this for you?
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Thread starter 5 years ago
#8
Sorry.... I didnt catch your post until this morning... thanks for the reply

In essence I suppose I am trying to understand how the current flowing through a resistor affects the amount of heat dissipated, and therefore the drop in voltage across the resistor..... specifically regarding resistors in series.

A battery with just a wire attached to each end will blow because there is almost zero resistance and therefore nothing to dissipate the energy as it flows through. Whereas with just one resistor the entire voltage is consumed. The coulombs exiting, even though they have no energy, continue on their way to the cathode due to the pressure exerted by fresh charge coming from the anode?

With an additional resistor added, the current may halve (for example) and therefore the amount of charge (per sec) passing through the first resistor has halved.
0
5 years ago
#9
(Original post by Science hopeful)
Sorry.... I didnt catch your post until this morning... thanks for the reply

In essence I suppose I am trying to understand how the current flowing through a resistor affects the amount of heat dissipated, and therefore the drop in voltage across the resistor..... specifically regarding resistors in series.

A battery with just a wire attached to each end will blow because there is almost zero resistance and therefore nothing to dissipate the energy as it flows through. Whereas with just one resistor the entire voltage is consumed. The coulombs exiting, even though they have no energy, continue on their way to the cathode due to the pressure exerted by fresh charge coming from the anode?

With an additional resistor added, the current may halve (for example) and therefore the amount of charge (per sec) passing through the first resistor has halved.
The wire will most likely melt before anything happens to the battery. An ideal dc voltage source will force any current to keep the voltage the same. If you short a battery then yes bad things happen, adding a resistor prevents this, but the resistor has to be the right size to control the current flow. If you add a small resistor to a large voltage source you are still going to have a large current flow in the network.

I think you should have a look at this link, it explains a lot.
0
5 years ago
#10
(Original post by Science hopeful)
In essence I suppose I am trying to understand how the current flowing through a resistor affects the amount of heat dissipated, and therefore the drop in voltage across the resistor..... specifically regarding resistors in series.

A battery with just a wire attached to each end will blow because there is almost zero resistance and therefore nothing to dissipate the energy as it flows through. Whereas with just one resistor the entire voltage is consumed. The coulombs exiting, even though they have no energy, continue on their way to the cathode due to the pressure exerted by fresh charge coming from the anode?

With an additional resistor added, the current may halve (for example) and therefore the amount of charge (per sec) passing through the first resistor has halved.
OK. I think I understand where you are having difficulty.

Apologies if you already know most of this, however, in your previous statements you stated a few things which indicated a few misconceptions. These need to be corrected before you can move forward:

1) Electric current is the movement of charged particles. In the case of current in a conductor, we are concerned with the movement of electrons exchanging between the atoms of the conductor.

2) Charged particles (electrons and protons) are carriers of the fundamental electromagnetic force. Like charges repel and unlike charges attract. Each and every electron will repel each and every other electron. Electric current is defined as a flow of charge (coulombs) per second.and called amperes.

3) In a battery (power source), chemical reactions create an excess of electrons (-ve charge) at the anode (-ve terminal) and a deficit of electrons (+ve charge) at the cathode (+ve terminal). When the terminals are connected externally by a conductor to create a circuit, the imbalance of charge between the terminals generates a repulsive force between the electrons at the -ve terminal (anode) and an attractive force for electrons at the +ve terminal (cathode).

4) Electrons in a conductor move by jumping between adjacent atoms. Movement of the electrons is like a ripple effect: Electrons pushed onto the conductor at the anode end, will repel the electrons already in the conductor and that repulsive force then ripples throughout the length of the conductor (rather like a Newtons cradle) such that at the other end, the net attractive force of the +ve cathode pulls electrons off the conductor.

5) As electrons move through the conductor, energy is used by increasing the vibrational motion of the conductor atoms. i.e. work is done, power (joules/second) is consumed.

6) The actual direction of electron flow in the conductor is opposite to the standard convention of current flow (+ve to -ve) which is used purely from a historical adoption.

7) Newtons law states that 'every action has an equal and opposite reaction'. This means, anything that impedes the passage of electrons through the conductor will generate an equal and opposite force which ripples back towards the anode.

8) Voltage is defined by joules per unit charge. i.e. it is analogous to pressure. The battery is able to supply a limited pressure force (voltage, magnitude dependent on the construction of the battery) which causes a current to flow in the conductor between the anode and the cathode.

THIS IS THE KEY BIT:

9) The forward voltage pressure at the anode will produce a current in the conductor which increases up to the point where it balances with the backwards ripple effect force of any resistance in the conductor. Current is therefore the ratio between battery voltage (forwards pressure) and resistance (reaction force) or V/R.

By definition, resistance is generated at all points throughout a conductor and is cumulative in effect, increasing with length.

10) Adding resistance will therefore reduce the current precisely because the forward and reaction forces must find a new equiibrium.

Beause equilibrium is dependent on the total resistance in the conduction path, the voltage pressure at the junction between the conductor and the cathode must always be zero. The voltage pressure therefore must reduce uniformly throughout the conductor and at an appropriate rate throughout any resistance in the path of the current.

You should also now be able to see that since current is dependent on the equiibrium of forces, the rate of energy consumption throughout the conductor/resistance is also dependent on that equiibrium. Hence why placing a lamp and resistance in series will share the energy consumption in the ratio of the resistances.

This is quite a long explanation, but I hope it goes some way in conceptualising what is actually happening and the relationshi between voltage, current, resistance, charge, energy and power in an electric circuit?
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Thread starter 5 years ago
#11
Firstly, can I just say thank you..... you put so much time in this reply and I mean this with the utmost sincerity, that the explanation is THE best I have ever seen or heard..... it should be posted somewhere highly visible on the internet.

(Original post by uberteknik)
1) Electric current is the movement of charged particles. In the case of current in a conductor, we are concerned with the movement of electrons exchanging between the atoms of the conductor.
UNDERSTOOD

(Original post by uberteknik)
2) Charged particles (electrons and protons) are carriers of the fundamental electromagnetic force. Like charges repel and unlike charges attract. Each and every electron will repel each and every other electron. Electric current is defined as a flow of charge (coulombs) per second.and called amperes.
UNDERSTOOD

(Original post by uberteknik)
3) In a battery (power source), chemical reactions create an excess of electrons (-ve charge) at the anode (-ve terminal) and a deficit of electrons (+ve charge) at the cathode (+ve terminal). When the terminals are connected externally by a conductor to create a circuit, the imbalance of charge between the terminals generates a repulsive force between the electrons at the -ve terminal (anode) and an attractive force for electrons at the +ve terminal (cathode).
UNDERSTOOD

(Original post by uberteknik)
4) Electrons in a conductor move by jumping between adjacent atoms. Movement of the electrons is like a ripple effect: Electrons pushed onto the conductor at the anode end, will repel the electrons already in the conductor and that repulsive force then ripples throughout the length of the conductor (rather like a Newtons cradle) such that at the other end, the net attractive force of the +ve cathode pulls electrons off the conductor.
(Original post by uberteknik)
5) As electrons move through the conductor, energy is used by increasing the vibrational motion of the conductor atoms. i.e. work is done, power (joules/second) is consumed.
UNDERSTOOD

(Original post by uberteknik)
6) The actual direction of electron flow in the conductor is opposite to the standard convention of current flow (+ve to -ve) which is used purely from a historical adoption.
UNDERSTOOD

(Original post by uberteknik)
7) Newtons law states that 'every action has an equal and opposite reaction'. This means, anything that impedes the passage of electrons through the conductor will generate an equal and opposite force which ripples back towards the anode.
UNDERSTOOD

(Original post by uberteknik)
8) Voltage is defined by joules per unit charge. i.e. it is analogous to pressure. The battery is able to supply a limited pressure force (voltage, magnitude dependent on the construction of the battery) which causes a current to flow in the conductor between the anode and the cathode.
UNDERSTOOD

(Original post by uberteknik)
THIS IS THE KEY BIT:
(Original post by uberteknik)

9) The forward voltage pressure at the anode will produce a current in the conductor which increases up to the point where it balances with the backwards ripple effect force of any resistance in the conductor. Current is therefore the ratio between battery voltage (forwards pressure) and resistance (reaction force) or V/R.

By definition, resistance is generated at all points throughout a conductor and is cumulative in effect, increasing with length.
In a basic circuit of a battery and wire, the wire functions as a very weak resistor. Nevertheless, using the previous examples, why is it that this set up can cause the battery to blow? Surely the resistor (the wire) should along its length dissipate enough energy to end up melting before the battery blows out? A voltmeter would still measure a for example 4V PD between terminals, meaning that 4 J/C had been expended..... so whats the pro?? Is this to do with the length of the wire used or that the sheer volume of current due to the low resistance makes this inevitable?

(Original post by uberteknik)
10) Adding resistance will therefore reduce the current precisely because the forward and reaction forces must find a new equiibrium.
UNDERSTOOD

(Original post by uberteknik)
Beause equilibrium is dependent on the total resistance in the conduction path, the voltage pressure at the junction between the conductor and the cathode must always be zero. The voltage pressure therefore must reduce uniformly throughout the conductor and at an appropriate rate throughout any resistance in the path of the current.
Is this because by the end of the conductor, there is now no PD between the anode and cathode? Surely the de-energised electrons would still be repelling each other?
0
5 years ago
#12
(Original post by Science hopeful)
Firstly, can I just say thank you..... you put so much time in this reply and I mean this with the utmost sincerity
Thank you for your kind words. Much appreciated. (Original post by Science hopeful)
In a basic circuit of a battery and wire, the wire functions as a very weak resistor. Nevertheless, using the previous examples, why is it that this set up can cause the battery to blow? Surely the resistor (the wire) should along its length dissipate enough energy to end up melting before the battery blows out? A voltmeter would still measure a for example 4V PD between terminals, meaning that 4 J/C had been expended..... so whats the pro?? Is this to do with the length of the wire used or that the sheer volume of current due to the low resistance makes this inevitable?
Not quite sure what you are getting at here?

I could read it in several ways:

a) failure caused by the battery overheating and a weak joint or internal wire fails before the external wire fuses. Alternately, the excesssive current throug a weak solder joint in the battery causes solder reflow which increases the internal resistance and hence most of the voltage is dropped across that new internal resistance.

b) The external short circuit allows the oxidation-reduction reaction rate to be uncontrolled and therefore runs away. The reaction is exothermic and thus the heat generated cannot be removed fast enough. This together with the gaseous products of the reaction in a confined volume causes the case to fracture or even explode. (usually, weak points are designed into the battery casing to allow any gas build up to safely vent before this happens.)

c) At the end of the battery life the reactants reach depletion. The rate of the chemical reaction slows significantly and the voltage pressure cannot be maintained. Under s/c load, the voltage between the anode and cathode therefore collapses.

d) If the voltmeter is the only thing connected to the battery across the anode and cathode, then since the voltmeter is purposely designed to present a very high load resistance, it will draw a very small current (microamps) from the battery. Hence the almost depleted reactants are nonetheless still able to supply that small current and the battery and the voltmeter will still register the full voltage (equilibrium). But as soon as the resistance falls (load increases) and therefore the current demanded increases beyond the rate of reaction capability, the supplied Joules/Coulomb must reduce and the voltmeter reading will fall.

(Original post by Science hopeful)
Is this because by the end of the conductor, there is now no PD between the anode and cathode? Surely the de-energised electrons would still be repelling each other?
At the cathode the electrons are now attracted to the receptor atoms. By definition therefore, the electrons have reached the end of their journey and the reaction of those atoms is complete.

Do not forget that the voltage measured between any two arbitrary points in a circuit, is a measure of the energy expended by the electrons in working against the resistance between those points. It does not mean the electrons cannot still supply energy further downstream. That only happens at the final resting place when the electron is captured by the cathode atoms.
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Thread starter 5 years ago
#13
Sorry it has taken so long to reply - I have been away with work and internetless!

Thank you for the above response it has clarified things nicely!
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