beginner help

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
i know this sounds like a dumb question, but does voltage increase if resistance increases? & why?(as opposed to current decreasing from resistance).
 

Dave

Joined Nov 17, 2003
6,970
The relationship between voltage and resistance, and current and resistance is defined by Ohm's Law (as I'm sure you know):

I = V/R

In the first instance, if we assume constant current, then if the resistance increases the voltage also increases (to maintain constant current).

In the second instance, if we assume constant voltage, then if the resistance increases then the current decreases (to maintain constant voltage).

As to why, think of Ohm's Law as defining the voltage, V, across a resistor, R, with current, I flowing through it. R is the resistance to current flow, so if we increase R the resistance to current increases and hence the current, I, through the resistor decreases for a constant voltage, V. For the increased resistance, if we increase the voltage, V, we can get the same current, I, as before we increased the resistance, i.e. constant current.

You can look at voltage, current and resistance on an atomic level to see how they influence each other, but the above is a conceptual description.

Dave
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
Oh ok i get it, so if we are keeping one value(1) constant and we change another value(2), than in order for the value(1) to stay constant, than we must change value(3) to oppose value(2).
 

Dave

Joined Nov 17, 2003
6,970
Oh ok i get it, so if we are keeping one value(1) constant and we change another value(2), than in order for the value(1) to stay constant, than we must change value(3) to oppose value(2).
Correct. Once you grasp this fundamental concept you can extrapolate the workings of Ohm's Law to any problem.

Dave
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
I also have a current confussion question(if you don't mind). you see, ive never really been to fimiliar with current, even though ive read about how it is the flow of the electrons.could this kind be considered the speed or rate of electron flow? Also, how exactly is related to current?
 

Dave

Joined Nov 17, 2003
6,970
Current is defined as the rate of change of charge, mathematically I = dQ/dt (i.e. the change in charge with time). So for every second the charge flowing through a particular point changes by a certain amount - this is the current. Therefore, your statement of rate of electron flow is pretty accurate.

Dave
 

Dave

Joined Nov 17, 2003
6,970
but what is a charge, and how does it change every second?
You are asking one of the fundamental questions of physics. The electric charge is a fundamental property of atoms which determines the electromagnetic interactions experienced by atoms - accept this as a given. You are aware that the atom is comprised of Protons, Neutrons and Electrons. From the perspective of electric charge, protons have a net positive charge, electrons have a net negative charge and neutrons have net neutral charge. You should also be aware that like charges repel and unlike charges attract.

The charge on one electron (and conversely one proton) is known as the elementary charge. All charge amounts are derived from this elementary (fundamental) charge, hence it is quantised (even divisible amounts in quarks, but this is way off-topic). The elementary charge is equal to 1.6e-19C - this is the charge on one electron. Because the elementary charge on a proton and an electron are equal and opposite, they will have the net effect of cancelling each other out.

Atoms are (often) comprised of several protons, neurons and electrons, therefore have a charge that is a multiple of the elementary charge. This is where it gets confusing: if the atom has the same number of protons as electrons then the net charge on the atom is neutral - seems logical since the net positive charge on the protons cancels with the net negative charge on the electrons. There is one important point to note - the protons are bound to the atoms neucleus and are therefore unable to move, however the electrons are not bound and will move under the influence of an applied electric field (you should be very confused at this point!!). This is very important: The applied electric field is the result of the applied voltage, so by applying a voltage you create an electric field which moves the electrons - this movement is your current which is responding the applied voltage. Remove the voltage and the electric field disappears and the current ceases to flow. This should answer your second question, how does charge flow (or change every second)? - Its because of the applied voltage.

Apologies that this is somewhat complicated, but I have tried to explain it best I can. Any questions, let me know and I'll try to clarify it.

Dave
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
Actually this helped tremendously, thanks. So practically, i see how current is the flow and result of the applied voltage(no V=no I).right? so should less voltage equal less current keeping resistance constant? That being said, and current being a cause of voltage, than current is an exact variation on voltage right?
 

Dave

Joined Nov 17, 2003
6,970
Actually this helped tremendously, thanks. So practically, i see how current is the flow and result of the applied voltage(no V=no I).right?
Correct.

so should less voltage equal less current keeping resistance constant?
Correct.

That being said, and current being a cause of voltage, than current is an exact variation on voltage right?
For the purposes of resistance, yes. The current-voltage relationship is slightly more complicated for inductors and capacitors, but do not concern yourself with this here.

Dave
 

Dave

Joined Nov 17, 2003
6,970
Another thing im having trouble with is using a shunt resister across a voltmeter to mesure current.(i dont get it!)
The shunt resistor is placed in series with the circuit which you are measuring the current through. The current flowing through the shunt resistor will cause a voltage drop across the resistor which you measure with the voltmeter. By using Ohm's Law: I = V/R (Current = Voltage drop across shunt resistor/Resistance of shunt resistor), you can derive the current flowing through the circuit.

Dave
 
You got that right Current times Resistance is Directly proportional to Voltage.

So When Current or Resistance increases so does voltage.
So when Current or Resistance decreases so does voltage.

(E)ven = (I) * (R)ememeber.....what Ohms law states.

You should always remember you can ONLY measure current in SERIES things go bad when you measure current in parallel. If you shunt a resistor in to measure current it is like throwing it into parallel so that means you might not be getting the correct value of current because your measuring the current that the resistor draws and not what the circuit is drawing.

Here is some information about what a shunt resistor is.
http://en.wikipedia.org/wiki/Shunt_(electrical)
 

Dave

Joined Nov 17, 2003
6,970
You should always remember you can ONLY measure current in SERIES things go bad when you measure current in parallel. If you shunt a resistor in to measure current it is like throwing it into parallel so that means you might not be getting the correct value of current because your measuring the current that the resistor draws and not what the circuit is drawing.
That is a good point. As a continuum from my previous post, it is correct to say that the shunt resistor will have an influence on the measurement for the quoted reason. Ideally, when you place your shunt/meter in series with the circuit you would like it to present a zero resistance (which is impossible for all practical purposes) in order for it to have no influence on the measurement. In the same way when you measure voltage you place your meter in parallel around the component(s) you measuring the voltage of, then you ideally want the meter to present an infinite resistance thus to prevent it drawing current and thus again influencing the measurement.

Dave
 

wireaddict

Joined Nov 1, 2006
133
Howdy, it's actually a shunt resistor connected across an ammeter or milliammeter to reduce the current through the meter thus increasing its range. It's been 30 years since I've studied instrumentation so I'm no expert here but if you have a 5 amp meter and place a shunt with 1/10 the resistance of the meter across it then it will take 50 amperes flowing through the meter and shunt for the meter to read full scale instead of 5 amps. Many times the current ratio is printed or stamped on larger shunts.
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
Thanks i get what all you guys are talking about, i read about it a week
ago. I now have another problem in electronics (as far as learning goes)
Im currently reading chapter 10 on DC network analysis of volume 1 DC,
and im having trouble with the branch current and mesh current method
(espesialy branch method).

Branch Current Method
&
Mesh Current Method

I get most of it, but my only problem is at the the end when it comes time to figure out the equations to come up
with the amount of current through the individual resistors(i dont know how to solve these particular equations!).
 

Dave

Joined Nov 17, 2003
6,970
Thanks i get what all you guys are talking about, i read about it a week
ago. I now have another problem in electronics (as far as learning goes)
Im currently reading chapter 10 on DC network analysis of volume 1 DC,
and im having trouble with the branch current and mesh current method
(espesialy branch method).

Branch Current Method
&
Mesh Current Method

I get most of it, but my only problem is at the the end when it comes time to figure out the equations to come up
with the amount of current through the individual resistors(i dont know how to solve these particular equations!).
Ok, lets deal with one method at a time: The Branch Current Method. Where exactly in the AAC tutorial are beginning to not understand it? What is posing you the problem?

Howdy, it's actually a shunt resistor connected across an ammeter or milliammeter to reduce the current through the meter thus increasing its range. It's been 30 years since I've studied instrumentation so I'm no expert here but if you have a 5 amp meter and place a shunt with 1/10 the resistance of the meter across it then it will take 50 amperes flowing through the meter and shunt for the meter to read full scale instead of 5 amps. Many times the current ratio is printed or stamped on larger shunts.
As it happens what you are saying is also correct, although using a shunt in the way I have previously described is also accurate. The situation you have described refers the shunt resistor placed in parallel with the coil in a Perminent Magnet Moving Coil (PMMC) ammeter. The PMMC meter experiences a full-scale deflection, and the shunt resistor connected in parallel is designed to increase its current measurement range (as you have described). The shunt resistance in this case is configured to ensure only a fraction of the current bypasses the coil and hence the full-scale deflection of the whole measurement system comprising the shunt resistor and PMMC meter is given by:

I(fs) = [(Rc + Rs)/Rs]*I(fs)coil

I guess this demonstrates there is an element of interpretation here.

Dave
 

Dave

Joined Nov 17, 2003
6,970
Apologies for the lateness of my reply, I was off-line yesterday.

Right your issues with the Branch Current Method:

I take it you are happy with the way the tutorial has used Kirchoff's Voltage and Current Laws, and Ohm's Law to derive the 3 equations:

-I1 + I2 - I3 = 0

-28 + 2*I2 + I1 = 0

-2*I2 + 7 - I3 = 0

If you have a problem here let me know.

The next step uses some simple rearrangement, placing the variables not expressed in I1, I2 or I3 on the right hand side of the equation, and ensuring that the left hand side of the equation is expressed in all three current variables I1, I2 and I3. If there is no current variable, then 0*I1, 0*I2 or 0*I3 (all of which are equal to zero) is used as a place holder for the simultaneous equation.

This gives:

-I1 + I2 - I3 = 0

4*I1 + 2*I2 + 0*I3 = 28

0*I1 - 2*I2 - I3 = -7

The three equations are then solved as simultaneous equations. For information on solving simultaneous equations refer to:

Volume 5 - Chapter 4 - Solving Simultaneous Equations

Any further questions on the above?

Dave
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
Oh... I see, i guess my problem was simultaneous equations. I can subsitute, but i just didn't know they were called "simultaneous equations", But i get it know.Thanks!
 
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