A voltage of 6 V moves through a resistor. This does not make sense??

[physicsishard]

Ok that is the question, the first part anyways. It does not make sense due to "VOLTAGE drop" (I think?) but I do not get why it doesn't resist CURRENT? I guess I am just confused about Ohm's Law in general. I get the formula of current equals voltage over resistance but I am just super confused. Thanks in advance

Edited November 20, 2016 by physicsishard

[Externet]

When voltage is applied to a resistor, current 'moves' through it, or flows. It flows in the amount the resistor value or resistance allows. Can be a little, can be a lot.

 

Think of a water pipe where the flow is restricted by an obstruction inside the pipe. The obstruction is the resistor. The litres per second is the current flowing, the water pressure applied to the pipe is the voltage.

A small obstruction will allow more current or flow of the water trough it.

 

For the Ohm's law formula, the current flowing is the pressure applied divided by the amount of obstruction.

[Sensei]

Ok that is the question, the first part anyways. It does not make sense due to "VOLTAGE drop" (I think?) but I do not get why it doesn't resist CURRENT? I guess I am just confused about Ohm's Law in general. I get the formula of current equals voltage over resistance but I am just super confused. Thanks in advance

 

That's why people should start from quantum physics, then continue to classic physics, like in this one example..

 

"Voltage drop" (U1-U0)=dU multiplied by q is energy E, which is energy that's lost by electrons during traveling through some element.

This energy is lost by electrons but still conserved, so element is heated, and have to radiate it away.

If element has high resistance, they lose a lot of energy, large voltage drop on element,

if element has low resistance, they lose little energy. small voltage drop on element.

To travel through element with high resistance there is needed large initial voltage.

(If element is superconductor, drop of voltage is so small that it's measured as near 0).

Say U=230 V, R=100 ohm, therefor I=U/R = 230 V/100 ohm = 2.3 A

But Q=I*t, so 2.3 A*1s = 2.3 C,

but Q is quantizied every e=1.602176565*10^-19 C,

so Q/e is quantity of electrons in wire flowing through it in 1 second period of time.

Q/e = 1.43555*10^19 electrons per second.

 

If we will have R=10 ohm, the same current I=2.3 A (and the same Q=2.3 C) could be achieved with just U=23 V.

 

There are elements like transformer

https://en.wikipedia.org/wiki/Transformer

which try to transform large voltage with small current,

to smaller voltage with larger current,

U₀*I₀=U₁*I₁

U₀>U₁

I₀<I₁

(in practice there are loses)

 

Again if you multiply both sides by time t

U0*I0*t = U1*I1*t

U0*Q0=U1*Q1

E0=E1

It's just energy conservation equation.

Or electrons:

U0*(Q0/e) = U1*(Q1/e)

Edited November 20, 2016 by Sensei

[Carrock]

That's why people should start from quantum physics, then continue to classic physics, like in this one example..

Your post seems to be pure classical physics, except for the mention of superconductivity.

 

From https://en.wikipedia.org/wiki/Cathode_ray

In 1897 British physicist J. J. Thomson showed the rays were composed of a previously unknown negatively charged particle, which was later named the electron.

The idea that matter is not infinitely divisible is surely a classical concept, originated in 5th century B.C.E. by Leucippus and his pupil Democritus, or earlier.

[Sensei]

Your post seems to be pure classical physics, except for the mention of superconductivity.

In pure classical physics, charge Q is infinitely dividable. In quantum physics Q is integer multiply of e.

 

Experiment in which e, elementary electric charge, has been calculated the first time, is oil drop experiment made by Robert A. Millikan and Harvey Fletcher.

https://en.wikipedia.org/wiki/Oil_drop_experiment

[Capiert]

Ok that is the question, the first part anyways. It does not make sense due to "VOLTAGE drop" (I think?) but I do not get why it doesn't resist CURRENT? I guess I am just confused about Ohm's Law in general. I get the formula of current equals voltage over resistance but I am just super confused. Thanks in advance

Hi

 

It looks like you are confused about the names. You've mixed them up, by the looks of it, which is an honest mistake when starting.

 

Voltage does NOT (really) flow thru the resistor.

 

CURRENT flows (instead),

which is called electrons per second.

 

Is that understandable?

 

Think of voltage like (water's) pressure

in a pipe.

 

& (a river's) current is the "flowing" water.

 

Voltage "drop" is like the water's pressure_drop.

 

The water comes out slower

at the end of a very long garden hose,

but stronger & faster from the tap (of the kitchen sink).

Or a bit like that.

 

The hose is like a (resistor) wire.

 

1 end (=output) has less pressure

when the water flows.

 

If the tap (faucet) is turned off,

then all the water inside

has the same (amount of) pressure.

 

"Resistance"

"resists" current;

 

NOT voltage_drop.

 

Voltage_drop

like pressure,

makes the current flow!

Edited November 21, 2016 by Capiert

[studiot]

Gentlemen,

Surely all this quantum / historical stuff is over the top for someone who is just starting to learn Ohm's Law?

 

 

Welcome to ScienceForums, physics is hard.

 

OK so here is a rough guide to electrics.

 

First and foremost you need to know that there are two viewpoints, just like with other subjects.

 

A practical man's overview and the underlying science.

 

It is good to get a bit of the practical overview before plunging into the underlying science.

Then you will have some familiar material to work with.

 

This is a bit like knowing that white powder A (sugar) will dissolve when you stir it into your tea, but white powder B (chalk) will not before studying the underlying scientific reasons for this.

 

So in electric circuits we have two kinds of elements or components.

 

Those that supply electrical energy (batteries, the mains and so on)

 

Those that dissipate or utilise electric energy (light bulbs, heaters and so on)

 

We measure the amount of electricity as the current in amperes or 'amps' for short.

 

The voltage or volts for short does double duty.

 

For sources of supply of electrical energy it is a measure of the driving force or electromotive force.

 

For dissipators and users of electricity it is the 'voltage drop' you asked about.

 

In order to get or force a particular dissipating element to pass a particular amount of current you have to impress a particular driving voltage across it from a electrical source.

We say the dissipating element 'drops' that amount of voltage.

 

Ohm's Law is the simplest relationship between voltage and current.

 

Notice I haven't (needed to) gone into the more detailed physics behind this.

 

Some diagrams would be helpful at this stage.

 

If you want to continue perhaps sensei or I can draw some.

[physicsishard]

 

Gentlemen,

Surely all this quantum / historical stuff is over the top for someone who is just starting to learn Ohm's Law?

 

 

Welcome to ScienceForums, physics is hard.

 

OK so here is a rough guide to electrics.

 

First and foremost you need to know that there are two viewpoints, just like with other subjects.

 

A practical man's overview and the underlying science.

 

It is good to get a bit of the practical overview before plunging into the underlying science.

Then you will have some familiar material to work with.

 

This is a bit like knowing that white powder A (sugar) will dissolve when you stir it into your tea, but white powder B (chalk) will not before studying the underlying scientific reasons for this.

 

So in electric circuits we have two kinds of elements or components.

 

Those that supply electrical energy (batteries, the mains and so on)

 

Those that dissipate or utilise electric energy (light bulbs, heaters and so on)

 

We measure the amount of electricity as the current in amperes or 'amps' for short.

 

The voltage or volts for short does double duty.

 

For sources of supply of electrical energy it is a measure of the driving force or electromotive force.

 

For dissipators and users of electricity it is the 'voltage drop' you asked about.

 

In order to get or force a particular dissipating element to pass a particular amount of current you have to impress a particular driving voltage across it from a electrical source.

We say the dissipating element 'drops' that amount of voltage.

 

Ohm's Law is the simplest relationship between voltage and current.

 

Notice I haven't (needed to) gone into the more detailed physics behind this.

 

Some diagrams would be helpful at this stage.

 

If you want to continue perhaps sensei or I can draw some.

 

Sorry I do not know how to quote only portion of your text. I am learning the basics of this Law so thank you. For example my professor asked "What types of circuit elements create a significant voltage drop" I am answering saying that it is when there are small amounts of conductor resistance. Does that make sense to you? And then "the type of circuit with zero or almost zero circuit drop" are circuits with zero resistance. I hope that makes sense and does not confuse you. Also sorry if my English is rough

[Sensei]

"What types of circuit elements create a significant voltage drop"

I am answering saying that it is when there are small amounts of conductor resistance. Does that make sense to you?

No, when you have small resistance, you have small drop of voltage, as kinetic energy lost by electron traveling through element is very small.

See my post #3..

 

Small resistance = large current.

Because resistance in inversely proportional.

 

[math]I=\frac{U}{R}[/math]

 

[math]Q=I*t[/math]

 

[math]Q=\frac{U}{R}*t[/math]

 

[math]E=Q*U[/math]

 

[math]E=\frac{U}{R}*t*U=\frac{U^2}{R}*t[/math]

 

[math]P=\frac{E}{t}[/math]

 

[math]P=\frac{\frac{U^2}{R}*t}{t}=\frac{U^2}{R}[/math]

 

(some of these equations are "hiding" quantization at quantum level)

 

ps. your teacher is asking about element type: resistor, capacitor, inductor, motor, lightbulb etc. etc.

 

And then "the type of circuit with zero or almost zero circuit drop" are circuits with zero resistance.

Circuit made with plain wires connecting negative and positive electrode on battery will blow up, damage wires and/or battery..

It's called short-cut/short circuit.

https://en.wikipedia.org/wiki/Short_circuit

Edited November 21, 2016 by Sensei

[studiot]

I said I would draw a diagram.

 

These are the first things you need to know.

 

I have built up the electrical circuit from Fig1 to Fig3, using an electrical source element (a battery) and an electrical dissipating element ( a light bulb).

 

The battery is said to have an 'electromotive force' or EMF of 10 volts.

This means that it maintains 10 volts between its terminals (shown as red wires) regardless of what is connected to it.

There is a potential difference or PD of 10 volts across its terminals.

Voltages are measured across an element.

 

The lightbulb in Fig1 is not connected to anything and because it has no electricity source it has zero voltage across its terminals.

 

There is no current flowing when there is no connection.

The electric source has a voltage across it.

The electric dissipating element has zero voltage across it.

 

Note that both the battery and the lightbulb have two terminals.

If we make one connection between one battery terminal and one lightbulb terminal as shown by a red wire in Fig2

There is no change.

 

We do not have a complete circuit so

There is no current flowing when there is only partial connection.

The electric source has a voltage across it.

The electric dissipating element has zero voltage across it.

 

If we now connect the remaining pair of terminals with another wire we have a complete loop or circuit.

This is shown in fig3.

 

Current now flows through both the battery and the lightbulb and they both have 10 volts across them.

 

Energy is now being transferred from the battery to the lightbulb and the current passes round the loop to return to the second terminal of the battery.

 

The energy transferred per second is called the power and we have our first equation.

 

Power = Current x Voltage = 1 x 10 = 10 watts or 10 Joules per second.

 

P = IV

 

So in 10 seconds 10 x10 Joules of energy will be transferred.

 

Curent = Power x time

 

I = PT

 

Now we come to Ohm's Law.

 

Firstly remember what I said backalong.

 

The battery maintains 10 volts.

 

This often causes confusion.

 

This means that the voltage is fixed at 10 volts.

The current is determined by the (electrical) size of the lightbulb.

The electrical size is called its resistance and is measured in Ohms.

 

Ohm's Law connects voltage, current and resistance by the equation

 

V = IR

 

The greater the resistance in Ohms, the lower the current that the lightbulb will draw.

 

Since I have fixed the voltage using the battery we have

 

10 = 1 x R

 

R = 10 Ohms.

 

 

To recap

Current moves (flows) through a resistor or other circuit element.

Voltage may be zero or have some value but does not move. It is always across some circuit element.

 

 

How are we doing?

Edited November 21, 2016 by studiot