Understanding Electricity

[Vastor]

Hey guys,

 

Firstly, there 3 chapter that I'm revised related (I thought) with this thread: Electricity, Electromagnetism, Electronics. This would be a long journey(If I have a lot question to ask)

 

So, I hope anyone can give me some intuition for me to solve my confusion.

 

let's get to the first doubt, it's about voltage in series circuit and the nature of voltage itself.

 

 

based on Textbook, Voltage = "The potential difference between two points is the work done in moving one coulomb from one point to another."

 

This give me a thought of that "I can put Voltage as a 'potential energy' in a circuit, but I wonder how that happen?(my first question)"

 

 

for the Potential difference in series circuits, I got this one diagram(from the textbook)

 

 

 

thus, in one of its "note" I found this,

Description:

The potential difference between two points in a a circuit is the potential drop from the point of higher potential to the point of lower potential.

Figure shows an example where the potential at p is the highest and the potential at r is the lowest.

 

 

 

After a while of thinking about these part, I eager to discover more about voltage, there something in my mind that is un-confirmed, yet. Based on the figure below. I'm putting an "imaginary Voltmeter" and put on a wild guess of its value, so can anyone confirm my thought on this?!

 

thnx.

[swansont]

In your last diagram, the "3V?" part on the left is, in fact 0V. There is no potential drop in that section of the circuit, because we ignore the resistance in the wires. Since there are two points at a potential of 3V, the difference between them has to be zero. The reading you'd get on the right is correct.

 

In reality, of course, there will be a small R and thus a small drop in V.

[Vastor]

In your last diagram, the "3V?" part on the left is, in fact 0V. There is no potential drop in that section of the circuit, because we ignore the resistance in the wires. Since there are two points at a potential of 3V, the difference between them has to be zero. The reading you'd get on the right is correct.

 

In reality, of course, there will be a small R and thus a small drop in V.

 

[Spyman]

Try to imagine the voltage potential as an height and electrons as bowling balls. The battery is an elevator lifting the bowling balls to an increased height of 3 meters and then they roll down back to the starting point. The lamp L1 has a heigth of 1.2 meters and the lamp L2 of 1.8 meters.

 

The black points are reference points for your measuring and don't have any heigths themselves and as swansont already has said the wires or cables represented by the black lines have such small heigth or voltage potential differences that they are usually neglected.

 

As such the 0V point is the starting point for the elevator or the ground level, the 1.8V point is 1.8 meters above the starting point or above the ground and the 3V point is 3 meters above the starting point or above the ground.

 

Between the 0V point and the 1.8V point there is a length of 1.8 meters or a voltage potential of 1.8 Volt, between the 1.8V point and the 3V point there is a length of 1.2 meters or a voltage potential of 1.2 Volts and between the 0V point and the 3V point there is a length of 3 meters or a voltage potential of 3 Volts.

 

Therefore there is no voltage potential difference between L1 and L2, in the analogy of heigths they are standing directly above/under each other.

 

In a water analogy the electrical current is the flow and voltage potential is pressure.

[Axioms]

The left side of the circuit is close to zero because there is hardly any change in potential difference (The resistance in the wires is really small). V=RI is the equation that defines how to work out a voltage in a simple circuit. Lets try work out what voltage is at the node on the left. V = (0 Ohm)(What ever the current is in the circuit).

 

From this you can see you have to have a component in the circuit that will cause the potential drop. Lets work out the voltage at the first resistor. V1=RI = (resistance of R1)(current in the circuit) [current will be the same through both resistors because the circuit is in series].

 

As you see in your first diagram the voltage across the first resistor and the second resistor added together is equal to the total of the circuit. You can see that because of the resistance to the flow of charge, there must be a difference in the amount of potential given to the system.

 

Think of it like this: you are pushing a block across a very slippery floor (the wire). You hardly have to push the block. You reach a patch that is really rough ground. You now have to push the block much harder to get it over this patch of ground (the resistor is the patch of ground and you having to push much harder is the voltage drop across the resistor). You get passed this patch of ground now and you are back on the slippery floor where you encounter hardly any effort to move the block.

 

 

[Joatmon]

Electromotive force (emf) is voltage that drives a current around a circuit. For example the voltage at the terminals of a battery.

Potential difference (pd) is voltage across a component that results from current flowing through the resistance of that component.

In a simple series circuit the sum of the pd's will equal the sum of the emf's. Since the pd's are developed with the opposite sign to the emf's if you work your way around the circuit the end result of adding pd's to emf's will equate to zero.

[Vastor]

Try to imagine the voltage potential as an height and electrons as bowling balls. The battery is an elevator lifting the bowling balls to an increased height of 3 meters and then they roll down back to the starting point. The lamp L1 has a heigth of 1.2 meters and the lamp L2 of 1.8 meters.

 

The black points are reference points for your measuring and don't have any heigths themselves and as swansont already has said the wires or cables represented by the black lines have such small heigth or voltage potential differences that they are usually neglected.

 

As such the 0V point is the starting point for the elevator or the ground level, the 1.8V point is 1.8 meters above the starting point or above the ground and the 3V point is 3 meters above the starting point or above the ground.

 

Between the 0V point and the 1.8V point there is a length of 1.8 meters or a voltage potential of 1.8 Volt, between the 1.8V point and the 3V point there is a length of 1.2 meters or a voltage potential of 1.2 Volts and between the 0V point and the 3V point there is a length of 3 meters or a voltage potential of 3 Volts.

 

Therefore there is no voltage potential difference between L1 and L2, in the analogy of heigths they are standing directly above/under each other.

 

In a water analogy the electrical current is the flow and voltage potential is pressure.

 

This explain what you said?

 

if my thought about voltage as "potential energy" is wrong, then it should be the "potential energy difference", based on your analogy and my current understanding of the voltage now.

 

 

The left side of the circuit is close to zero because there is hardly any change in potential difference (The resistance in the wires is really small). V=RI is the equation that defines how to work out a voltage in a simple circuit. Lets try work out what voltage is at the node on the left. V = (0 Ohm)(What ever the current is in the circuit).

 

From this you can see you have to have a component in the circuit that will cause the potential drop. Lets work out the voltage at the first resistor. V1=RI = (resistance of R1)(current in the circuit) [current will be the same through both resistors because the circuit is in series].

 

As you see in your first diagram the voltage across the first resistor and the second resistor added together is equal to the total of the circuit. You can see that because of the resistance to the flow of charge, there must be a difference in the amount of potential given to the system.

 

Think of it like this: you are pushing a block across a very slippery floor (the wire). You hardly have to push the block. You reach a patch that is really rough ground. You now have to push the block much harder to get it over this patch of ground (the resistor is the patch of ground and you having to push much harder is the voltage drop across the resistor). You get passed this patch of ground now and you are back on the slippery floor where you encounter hardly any effort to move the block.

 

when you said "there must be a difference in the amount of potential given to the system", so basically voltage is like a resultant differences that cause something to happen? I mean, like resultant force, if it's not 0 N, then the object will accelerate, or something like that?!

 

 

Electromotive force (emf) is voltage that drives a current around a circuit. For example the voltage at the terminals of a battery.

Potential difference (pd) is voltage across a component that results from current flowing through the resistance of that component.

In a simple series circuit the sum of the pd's will equal the sum of the emf's. Since the pd's are developed with the opposite sign to the emf's if you work your way around the circuit the end result of adding pd's to emf's will equate to zero.

 

Except for the fact that I learned electromotive force as :

 

e.m.f. = Potential Difference(Volt) + Drop in potential difference due to internal resistance(Resistance in the battery)

so basically, even though both things are the same, the textbook claim that "voltage is e.m.f., but e.m.f. is not voltage"

[Joatmon]

Except for the fact that I learned electromotive force as :

 

e.m.f. = Potential Difference(Volt) + Drop in potential difference due to internal resistance(Resistance in the battery)

so basically, even though both things are the same, the textbook claim that "voltage is e.m.f., but e.m.f. is not voltage"

 

Not quite. In simple calculations the battery is usually assumed to have no internal resistance. In this case the terminal voltage is a measurement of emf, the force that drives the current around the circuit.

When looking at the situation more closely you take the internal resistance of the battery into account. This means that, unless the battery is not supplying current, what you measure at the battery terminals is the battery emf minus volt drop(pd) across the internal resistance. In other words you include the battery internal resistance as one of the resistors in the circuit.

Edited March 23, 2012 by Joatmon

[Spyman]

This explain what you said?

(image snipped)

if my thought about voltage as "potential energy" is wrong, then it should be the "potential energy difference", based on your analogy and my current understanding of the voltage now.

Yes, it seems as if you got the general ide.

[Vastor]

Hey guys,

 

got 3 more question here,

 

1. if the charge model below is the right one, how does 2 negative charge repel if both are pulling?

 

 

 

 

2. From the textbook:

 

"Household electrical appliances that work on the heating effect of current are usually marked with voltage and power ratings. A typical filament bulb shown in Photograph 2.9 is marked 240V 60W. This means that the bulb will consume 60J of electrical energy every second if it is connected to 240V supply."

 

what happen if it's supplied with more or less than 240V? as far as I thought.

 

[math] E = VIt [/math]

 

and

 

[math] P = VI [/math]

 

thus, with constant Power, Voltage varies inversely to Current.

 

so, even though how different voltage is, it's only affect the current but not amount of energy, doesn't it?!

 

[math] E = Pt [/math]

 

 

 

3. Actually, I'm not quite understand of resistance yet, so hope anyone can help me on this question and give me a little insight about resistance itself.

 

based on the circuit below:

 

 

(i) What will happen to the bulb if the resistance of the rheostat is reduced?

 

(ii) Explain your answer in (a) (i).

 

for first question, as far as I learned about resistance, I just understand it as an invariant for Voltage that directly varies to Current (in Ohmic conductor)

 

[math] V \propto I [/math] or [math] \frac{V}{I} = R [/math]

 

thus, if R is reduced, the Voltage decreased OR the Current increased.

 

 

thank you for everyone's help!

[Joatmon]

Starting with "1". As you obviously know, the rule is like charges repel, unlike charges attract. The arrows are not directions of force. They represent the direction that "conventional current" would flow when driven by a voltage difference across a component. In the history of this assumption it wasn't known which direction the mysterious current went. A convention was made that whatever it was it would flow from positive to negative. Later, it was found that electric current was a flow of electrons that, in fact went in the other direction! So electron flow is from negative to positive! Later still it was realised that "holes" (places where electrons could fit as they travel) move in the direction of conventional current. These "holes" are known as positive charge carriers.

So conventional current and positive charge carriers move from positive to negative.

Electrons flow from negative to positive.

 

Looking at "2". If you have a filament lamp rated at 240V 60W it will provide (or take from the supply) 60W of power if connected to a 240V supply. If you increase the voltage you will also increase the current so the power will increase. Assuming the resistance of the lamp doesn't change doubling the voltage will increase the power four times. Because the lamp will be much hotter the resistance will change to some extent, but this is usually ignored in basic calculations. Halving the applied voltage will result in using a quarter of the power by the same argument. By Ohm's law the current through a fixed resistance (such as your lamp) is proportional to the applied voltage (other factors such as temperature remaining constant).

 

Looking at "3". The total resistance of the circuit is the sum of the resistance of the lamp (which is usually assumed to remain constant) plus the resistance of the rheostat. Rt=R1+R2.

Thus if you reduce the resistance of the rheostat you will reduce the total resistance of the circuit.

The current around the circuit will increase. The power given to the lamp will increase P=(I^2)*R( R=resistance of lamp). The lamp will glow more brightly.

The voltage across the lamp will increase. V=I*R (resistance of lamp). There is another way of determining the voltages across lamp and rheostat known as the voltage divider principle. The voltages across each will be in proportion to their resistances and add to give the supply voltage.

Edited March 25, 2012 by Joatmon

[Spyman]

3. Actually, I'm not quite understand of resistance yet, so hope anyone can help me on this question and give me a little insight about resistance itself.

 

based on the circuit below:

(Image snipped)

(i) What will happen to the bulb if the resistance of the rheostat is reduced?

A water analogy can be useful in understanding resistance, water pressure is voltage potential and water flow is electrical current. The battery is a pump increasing pressure, the rheostat is a valve choking the flow and the lightbulb is a waterwheel powered by the flow of water.

 

Unless the circuit is fatal for the battery and rapidly exhausting it, it will keep the voltage fairly constant independent of the load. In the analogy the pump is assumed to keep a constant pressure before the valve independent of it is fully open or closed.

 

The rheostat is a variable resistance throttling the electrical current through it depending on the voltage drop over it. In the analogy the valve strangles the flow of water by closing the hole water is passing through, causing a pressure drop over it.

 

The lightbulb is also a resistor where the electrical current are choked by its resistance depending on applied voltage. In the analogy the waterwheel will oppose the movement and cause the water flow to slow down, a higher pressure will cause a higher flow.

 

Therefore by reducing the resistance you are opening the valve and a larger hole will allow more water to pass. Since more water are allowed to reach the intake of the waterwheel, the pressure pushing water through it will increase, which will cause a greater flow. The waterwheel will start to spin faster or the lightbulb will shine brighter.

[Vastor]

22. The electromotive force and the internal resistance of the dry cell in the circuit below are 2V and 0.1(Ohm) respectively.

 

 

What is the reading of the voltmeter when the switch is closed if the resistor is replaced with a larger resistance?

 

a. 0V

b. 2V

c. less than 2V

d. more than 2V

 

I answered "d", but the given answer is "c".

 

I thought that more resistance, more 'joule per coulomb' needed to push the charge, right? V = R*I, right?

 

 

35. Two bulbs rated 240V 60W and 240V 100W. are connected in series and then in parallel to the mains supply. Which of the following correctly describes the brightness of the 60W bulb compared with the 100 W bulb?

 

Brightness (In series : In parallel)

A. 60W > 100W : 60W > 100W

B. 60W > 100W : 60W < 100W

C. 60W < 100W : 60W < 100W

D. 60W < 100W : 60W > 100W

 

the answer is B, which mean 60W bulb is brighter in series circuit but dimmer in parallel circuit. I try using the Ohm's Law relationship (R=V/I) but failed to recognized which one is needed to determine the brightness, thought so, I got a presumption where the brightness depend on energy it receive (per second). But then I got messed up.

 

help? thanks

[swansont]

For 22, if the EMF is 2 V and there is internal resistance, is it possible for there to be more than 2V (or =2V) across the rest of the circuit?

 

For 35, the brightness does depend on power, which is P = IV = I^2R. You also know that V=IR. Bulbs are designed to have their advertised brightness for the full voltage drop; that should give you relative sizes of R. In series the current is the same, so the one with the smaller R will have a smaller power.

[Joatmon]

Q22 You might also like to have another look at #8 and consider how V in your circuit compares with the cell terminal voltage.

[studiot]

Except for the fact that I learned electromotive force as :

 

e.m.f. = Potential Difference(Volt) + Drop in potential difference due to internal resistance(Resistance in the battery)

so basically, even though both things are the same, the textbook claim that "voltage is e.m.f., but e.m.f. is not voltage"

 

Both things are not the same they are just measured in the same units - volts.

 

It works like this.

 

I have an 18inch ruler and a 6 inch ruler.

 

That is the length of ruler A is 18 inches and the length of ruler B is 6 inches.

 

But the length difference between the two rulers is 12 inches.

 

Both length and length difference are measured in inches.

 

So it is with EMF and Potential Difference which are both measured in volts.

 

If it is any further help, neither are any form of energy.

However an EMF is a way of introducing electrical energy into a circuit whereas a PD is a way of dissipating electrical energy out of a circuit as heat.

[Vastor]

For 22, if the EMF is 2 V and there is internal resistance, is it possible for there to be more than 2V (or =2V) across the rest of the circuit?

 

For 35, the brightness does depend on power, which is P = IV = I^2R. You also know that V=IR. Bulbs are designed to have their advertised brightness for the full voltage drop; that should give you relative sizes of R. In series the current is the same, so the one with the smaller R will have a smaller power.

 

great, I got that. though for 35, I got another question, does the arrangement of the bulb (in series circuit) affect it brightness? I mean:

 

either this, or the other way (the bulb exchange places). could it affect the brightness?

based on your answer to 35, I don't get how my teacher claim there are brightness difference when changing the position due to the voltage amount would be depend on resistance anyway.

 

btw,(for 35 too) in parallel circuit, the voltage is the same. so, any bulb that has higher current would create more power, right?

 

 

thanks for others who replied too.

Edited July 15, 2012 by Vastor

[Vastor]

Hey everyone,

 

4.(a) A student finds that the brightness of three bulbs is different in series circuit and in parallel circuit, although the three bulbs and the new dry cells used are identical. Each bulb is lablelled 3V 0.3A. Figures 3 and 4 show the brightness of the bulbs in series circuit and in parallel circuit respectively.

 

(iii) Compare the brightness of the bulbs in each circuit.

Relate the brightness of the of the bulbs to the potential difference and the current flow for the bulb s in series and parallel circuits.

 

The confusion is based on the given answer:

 

-> The total potential difference in a series circuit is the sum of the potential differences across each bulb.

-> The current in a complete circuit of a series circuit is the same as the current flow through each bulb.

-> Therefore, the brightness of each bulb in the series circuit is the same. (makes sense!)

 

-> The potential difference across each bulb in a parallel circuit is the same as the potential difference across the cell.

-> The total current in the parallel circuit is the sum of the currents that pass through each bulb.

-> The brightness of the bulbs in the parallel circuit is more than that in the series circuit. (WTHell?!)

 

I'm informed that Brightness = Power, right? so let analyse something here...

 

 

 

let's give the voltage of the cell 3V, internal resistance ignored.

 

The information that we have:-

Bulb:-

voltage: 3V

current: 0.3A

resistance: 10'Ohm' (R = V/I)

power : 0.9W (P=VI)

So, it's mean that the bulb will consume 0.9joule of electrical energy every second if it is connected to a 3V supply.

 

Series circuit:-

voltage: sum of bulb voltage.

current: I of bulb = I of another bulb.

 

Parallel Circuit:-

voltage: V of bulb = V of another bulb.

current: sum of bulb current.

 

Given that P=VI, now, doesn't it's weird? I mean:

 

For series circuit: The power of each bulb

[math] P = (1/3)V * I [/math]

1/3 due to the sum of bulb voltage = total voltage.

 

For parallel circuit: the power of each bulb

[math] P = V * (1/3)I [/math]

1/3 due to the sum of bulb current = total current.

 

Conclusion: The power should be equal, thus the brightness are just the same.

 

 

I'm not done yet, it's posibble to "make" the bulb on parallel circuit glow brighter.

 

based on Ohm's law, the resistance should be constant. So, voltage is directly proportional to current. Thus increasing the voltage would increase the current.

 

and for P=VI.This create great amount of power difference between series(0.1W) and parallel(0.9W). but let's see the calculation:

 

Series

[math] P = 1V * 0.1A(10R=\frac{1V}{I}) = 0.1W [/math]

 

Parallel

[math] P = 3V * 0.3A = 0.9W [/math]

 

The question is, where does the current come from? Induced from the Ohm's law equation? It's not make sense, I mean there given information that told both circuit use identical bulb(equal resistance) and identical cell(equal voltage supply). doesn't that mean total current should be equal too?

 

hope anyone can give some insight.

 

 

btw, theres one more.

(b) An electrical circuit can be connected in series and in parallel. Given the advantage and disadvantage of a parallel circuit.

 

given answer:-

Advantage = The circuit can still be used if one of the bulbs is blown.

Disadvantage = The brightness of the bulbs is the same as the brightness of a bulb in a single circuit.

 

No problem here, just want to share for the amusing "Disadvantage" answer... ;P

Edited July 16, 2012 by Vastor

[swansont]

The question is, where does the current come from? Induced from the Ohm's law equation? It's not make sense, I mean there given information that told both circuit use identical bulb(equal resistance) and identical cell(equal voltage supply). doesn't that mean total current should be equal too?

Batteries, or any other voltage source, have a fixed voltage but will supply as much current as needed to satisfy Ohm's law. (Batteries supplying more current will drain faster, since the amount of energy stored in them is limited.)

[Spyman]

The question is, where does the current come from? Induced from the Ohm's law equation? It's not make sense, I mean there given information that told both circuit use identical bulb(equal resistance) and identical cell(equal voltage supply). doesn't that mean total current should be equal too?

As swansont says the battery is the supplier and it encounters different loads, 3×10=30 ohms in series vs 10/3=3.33 ohms in parallel.

 

Ohm's law tells us how much current a certain voltage is able to squeeze through a load but not how much the source can sustain before it depletes.

 

If the battery is able it will deliver 3/30=0.1 amps with bulbs in series and 3/3.33=0.9 amps when in parallel.

 

When the battery drains its output voltage will drop in accordance with the current such that Ohm's law is always fulfilled for the load.

[Vastor]

given voltage at transmission wire is 24V a.c., given the resistance of the wire is 30"Ohm".

 

so based on V/R = I, the current is 0.8A.

 

the power loss due to heating?

P = I^2 * R

P = 0.8^2 * 30 = 19.2W, right?

 

above are given question & answer from past year official exam.

 

 

now my question, if we set voltage at 48V a.c., resistance is still 30"Ohm".

 

so I = 1.6A

 

the power loss due to heating = 1.6^2 * 30 = 76.8W.

 

so, how "Increasing voltage reduce power loss in transmission" statement make sense?

[Joatmon]

This is homework, but I think I can say your aren't considering the load that the transmission lines are feeding. Think of the effect of keeping the power developed in the load the same for the different voltages. Then calculate voltage and current for both cases. Then calculate power loss in the transmission lines. You should find the higher voltage (with it's smaller current) should give less power loss in the transmission lines.

If you still have a problem then show some working.