# Circuit polarity in an RLC circuit ?????

Discussion in 'General Electronics Chat' started by wdflannery, Dec 24, 2007.

1. ### wdflannery Thread Starter New Member

Dec 24, 2007
4
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Let's work our way up to an RLC circuit. I'm confused at every step.

1. We have a circuit with a battery and a resistor. The battery symbol is positioned on the page + end up, and we have a clockwise loop starting at the top of the battery and going through a resistor, and returning to the negative terminal of the battery.

We agree that clockwise current is positive.

I have just read the page on the polarity for such a circuit on this site, and the resistor is labelled so that positive current flows from the + terminal to the - terminal.

Thus if Vb is the battery voltage and Vr is the resistor voltage, when current is flowing Vb = -Vr and Vb + Vr = 0 so Kirchoff's law is good.

What's confusing - the battery and resistor have different orientations (- terminal to + terminal) vis a vis current flow. Why?

2. Ok, what if we add a capacitor. How do we label the terminals? I did not see a page for this. Is there one?

If we orient it like the resistor, then a positive current does give the capacitor a positive charge. But, this positive charge on the capacitor is oriented opposite to the battery. So, we have voltage sources with different orientations.

If we orient the capacitor the other way, it takes negative current to charge it positively. dv/dt = -i/C ???

3. Add an inductor. Suppose we label it like the resistor. When it acts as a current source sending out positive current, then the positive current is coming from the - end !!!!!

Also, increasing current produces a positive voltage that opposes the increase, so v = Ldi/dt ???

Really I need a good tutorial on this business. Is there such a page on the net? In a book? I thought I understood it, but on closer inspection, I don't !!!!

2. ### Eyas Member

Dec 23, 2007
10
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Good day wdflannery,

I think if we start with the first point the rest will be very easy.
I have some rules in my head up here and I use it always in my cources.

FIRST: forget about the current through the battry.

SECOND: the current ALWAYS moves from the high voltage to the low voltage. Think about it like the water.

Now in the first point, we have a battery and a resistance connected to a a battery that has a + up and - down. let us apply KVL, we are going counterclockwise starting from the - side of the battery. from - side to + side the voltage increases "thus +Vb" then from the + side through the resistance to the - side the voltage decreases "thus -Vr" and the sum is zero. thus
Vb-Vr=0 ====> Vb=Vr

you can think about it as if you dealing with money, you gain and you lose.

OR

we can stick with the voltage polarity in the battery, the upper side of the resistance has a voltage +Vb and the lower side has a zero "as the ground" [OR simply the voltage across the resistance is Vb] so the current is as follows:
i=Vb/R.
============
Now the second and the third point is somehow tricky bu it is also simple

if we dealing with DC circuit that is the voltage is constant, then, after a long time the capacitor is replaced by an open circuit and the inductor by a short circuit and the result will be a simple resistive network that can be solved using the first part above.
HOWEVER, at the begining the capacitor and the inductor will start to charge and during this the current and the voltage is varying with time which you can find some text book that deal with this "may be I can post some of them for you" because it is somehow lenghty.

In dealing with AC circuit which has the voltage like the one in our house, the capacitor and the inductor can be replaced by a load with the following values:
C=1/(jwC)
L=jwL
and dealing with them as if they are resistors where all DC rules are applicable.

I will try to find some sources that can explain this effictively.

I am sorry if I fail to explain this and to use so many words

THANK YOU FOR TAKING THE TIME TO READ THIS LOOOOOOOOOOOOOOOOOOOONG REPLY AND I AM LOOKING FORWARD TO HEAR FROM YOU SOOOOOOOOON

GOOD LUCK

3. ### Eyas Member

Dec 23, 2007
10
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Oh I am sorry

resistors and inductors have no polarity but we label the polarities according to the direction of current across them or {+ side in the terminal that has higher voltage than the other terminal which is -}

capacitors have polarity and are labeled with + at the high voltage and - in the other terminal.

when replacing
C -> 1/(jwC)
L -> jwL
then the polarity is exactly the same as if they were resistors.

sorry again

4. ### Eyas Member

Dec 23, 2007
10
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Here is a website that can help you understand the basics of electrical engineering and it discuss the phesics beyond it also.

http://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/EEIndex.html

another thing is that usually the current through the battery is going out of + and in the - not like the other elements. there are cases where current in the battery is going in the + and out of the - where the battery is being charged.

5. ### wdflannery Thread Starter New Member

Dec 24, 2007
4
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Thanks Abu, I finally think I've got it straight in my mind. To do it, I had to distinguish beween 'applied' voltage and 'across' voltage. An applied voltage wants to push positive current through a device, a positive 'across' voltage goes from low voltage to high voltage in the direction of positive current. Thus, applied voltage = - across voltage !

Then the device equations are
Vr,across = -IR for a resistor
Vc,across = -i(1/C) integral (I) for a capacitor
Vi, applied = L di/dt for an inductor
a battery has a positive across voltage in the usual orientation

Kirchoff's voltage law sums across voltages.

I think I better quit for the day before I find a flaw in the above !

6. ### Eyas Member

Dec 23, 2007
10
0
there is no difference between applied and across voltages

the voltage across "anything" is difference between the voltage in one side and the voltage in the other side.

the applied voltage is the voltage you apply as a battery or any voltage supply. in our example the applied voltage is the voltage of the battery.

let me explain something here:

*the current always "always" moves from HIGH voltage to LOW voltage ALWAYS. this is the real direction of current.

**the direction of current and the voltage are two faces for the same coin. THUS, if you know the voltage across [voltage in one side - voltage in the other side] then you know the current and its direction. let us take an example

OOOOO- V +OOOOOOOOOOOOOOOOOOOOOOOO- V +OOOOOOI
given ----ww---- then we know the current is ----ww-------<<--

and

OOOOO+ V -OOOOOOOOOOOOOOOOOOOOOOOO+ V -OOOOOOI
given ----ww---- then we know the current is ----ww------->>--

***in solving most of the cases where you are asked to solve a circuit, what most of electrical engineers do is that they assume [PURE ASSUMPTION] the direction of the current and based on that they label the voltages, so the positive side of the load[R, C, or L] if the side the current enter and the negative side is the one that the current leave. then, they solve the circuit to find the values of the voltage and current.

NOW if the current they got is negative, it means that the direction they assumed is WRONG and the real direction is the opposite but the value is the same. so because the direction is opposite the voltage they labeled before is also opposite according to the RULE we discussed in the point above(**).

I want you to try the resistor and battery example again and try more than one time. each with different way. assume one direction and then solve and then assume the opposite direction and then solve.
the values must be the same but the sign will be different.

GOOD LUCK and tell me what you get.
I want to hear from you soon.

7. ### wdflannery Thread Starter New Member

Dec 24, 2007
4
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the current always "always" moves from HIGH voltage to LOW voltage ALWAYS

Things only start to get interesting when you have an inductor. The differential equation for an inductor is v = L di/dt, so the polarity of the voltage on an inductor is positive if i is increasing and negative if i is decreasing, regardless of the direction of i. So if positive current is decreasing, the voltage is negative and current is flowing from low to high.

This is one reason why I think it is advisable to think of an inductor as a current source. When the voltage and current direction are the same, the source is energized, when they are not the same, the source is depleted. Also, I have started thinking of the voltage on an inductor as always being 'applied', rather than 'generated' by the inductor. The inductor wants to generate current, and will do so unimpeded until an opposing voltage is applied to the inductor.

8. ### BlueDevry Member

Dec 26, 2007
22
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How does the negative voltage affect the real voltage source ?

9. ### wdflannery Thread Starter New Member

Dec 24, 2007
4
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Hmm... let's see. Suppose you have a circuit with a battery, a switch, an inductor, and a resistor. You close the switch, the voltage is applied to the inductor, positive current starts to flow clockwise,current increases, and the voltage drops across the inductor and the resistor. When the current reaches a steady state there is no voltage across the inductor and the whole voltage drop appears across the resistor.

Now instantaneiosly replace the battery with a wire. Current continues to flow, the voltage drop still appears across the resistor, however, now it is applied to the inductor, so the inductor current decreases, ultimately to 0.

So, you can consider, I suppose, the inductor as becoming a voltage source. But that doesn't feel right to me. It seems clearer to me if you think of the inductor as a current source. Then the current produces the opposing voltage in the resistor, which collapses the magnetic field in the inductor and decreases current flow to 0.

..... I think......

10. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Indeed! The collapsing field around the inductor produces EMF, just as would in any relative movement between field and conductor.

Voltage potential across the inductor is the derivative of current divided by the derivative of time. This can lead to tremendous voltage potential when opening a switch in series with the inductor. The phenomenon is called "inductive kick." This is why we use "snubber" diodes to protect circuits.

For an example of harnessing the inductive kick, run an internet search on "boost converters" and "buck converters."

11. ### kkazem Active Member

Jul 23, 2009
160
26
Hi all,

I hope this helps. Here are a couple of basic facts of ckt analysis:

1) In you circuit described above with only a battery and a resistor; it's a no-brainer as the voltage across the resistor is the battery voltage. This is the case in any circuit with only 2-terminals. And it's the long-time convention of electrical engineers like myself (30+ yrs experience) to have the current pointing out of a voltage source, like the battery. Although this is technically incorrect from a physics point of view of electron flow which states that the real electron current in a battery comes out of the negative terminal. For engineering circuit analysis, simply disregard this and pretend that the current comes out the positive terminal of any voltage source and you'll be fine.

2) For loads (non-sources) like resistors, caps and inductors, lets start simple with a resistor. The side of the resistor that the arrowhead of the current flows into is always the positive side and this is the same for inductors and caps being driven by a source (storing-up energy, rather than releasing it). For an inductor, if the source that causes the energy to store in the inductor is removed or switched off instantaniously, the same current level (only at the instant it's switched, then it decreases from there) and direction continues to flow out of the inductor, which now becomes a source instead of a load. The only differences are that the polarity across the inductor switches to the opposite direction at the instant the inductor is no longer storing energy and continues that way until 1 of 2 things happen. Either the current in the inductor reduces to zero, at which time it has no voltage across it nor any current flowing thru it. The other difference is that when the source switches off and the inductor becomes a source, as someone above correctly stated, the time rate of change of the inductor current will be negative, meaning that the current will be getting smaller, whereas when it's taking or storing energy, the time rate of current change in the inductor will be increasing.

Regards,
Kamran K.
kkazem