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Circuits - Incredibly Displeasing

  1. Apr 2, 2012 #1
    I'm having an unpleasant time studying circuits. I feel like they are much harder than the rest of Electromagnetic concepts in Calculus-based Physics II. I can understand Gauss' law and other concepts but circuits make my stomach cringe. It is the type of thing that begs the "do this and this and you get this." and the "remember this and this because I said so."

    When I study physics, I always have to know WHY and HOW and conceptualize everything as much as possible. But with circuits, I'm forced to "shut-up and calculate."

    Does anyone else feel this way?
  2. jcsd
  3. Apr 3, 2012 #2


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    Maybe it would help if you gave us some concrete examples of concepts or methods that you felt you were just being forced to accept and memorize, rather than being provided with the underlying rationale for them?

    There is definitely some physics involved. Kirchoff's Voltage Law (KVL, also known as the loop rule) is basically just the conservation of energy. It says that the sum of the potential differences around a closed loop in a circuit is zero. Charges can't have a net gain or loss of potential energy around a loop, else energy would not be conserved.

    Kirchoff's Current Law (KCL, also known as the junction rule) says that the sum of currents going into a junction (or node) is zero. In other words, the total current entering the node must be equal to the total current leaving the node. This is just an expression of the conservation of charge. If more charge were entering a node than leaving it, then charges would have to be disappearing into thin air. Similarly, if more charge were leaving a node than entering it, then charges would have to be magically created out of thin air.

    Both KVL and KCL are limiting approximations to what Maxwell's equations say would happen. The approximations are valid when certain conditions are met.

    Ohm's law shouldn't be taken as a strict "law", but rather a result that has been shown experimentally to be true (or approximately true) for some devices/materials.

    Perhaps you have trouble with so-called "idealized" circuit components such as an "ideal current source" that outputs a constant current no matter what? These are not meant to correspond to any sort of realistic devices or components. They obey the rules that they do simply because we stipulate that they do so. (In other words, they obey those rules by definition). They are useful pedagogically in order to learn circuit theory without overly complicating things by having to adopt a more realistic model for the behaviour of a device. That comes later.

    If you wanted to implement a constant current source or a constant voltage source in real life, it would be non-trivial and require some active electronics, not just passive components.

    That's what I can think of off the top of my head.

    EDIT: Also, sign conventions can take a bit of getting used to. But they are just that: conventions. We assume a polarity for the voltage across a resistor in a circuit, and then we assume that the current flows across the resistor in the direction from high potential to low. If we then get a "negative" current as an answer from doing our circuit analysis, all it means is that the actual direction of the current across the resistor (and indeed the voltage across the resistor) was in the opposite direction from what was assumed.
    Last edited: Apr 3, 2012
  4. Apr 3, 2012 #3
    I personally am a fan of the "water in pipes" model of a circuit, whereby electricity is water, current is water flow, wires are pipes, batteries are like pumps, resistors are like waterwheels or turbines, diodes are like one-way valves, and transistors are like pressure-gated valves. (You can sort of extend this to capacitors--capacitors are roughly like reservoirs or tanks. Maybe inductors are roughly like siphons? The analogy starts to break down with these "AC" circuit elements.)

    These help a lot with visualizing certain aspects of circuits. For example, you can make sense of the parallel/series resistor rules. Suppose you have a circuit where initially, a battery is hooked up to a big resistor, and later we add a second small resistor in parallel with the first.

    The initial situation is like pumping water through a turbine which takes a LOT of force to drive. The turbine's back-pressure causes the pump to only drive a small volume of water per unit time.

    The later situation corresponds to inserting an easily-driven turbine in parallel with the first turbine. Most of the water will be diverted through this relatively low back-pressure channel. So the volume of water the pump drives per unit time increases.

    Whereas adding the second turbine in series with the first would just increase the back pressure on the pump, lowering the amount of water the pump drives per unit time.

    You can use this reasoning to make sense of basic circuits (Kirchoff's laws) and even logic circuits with transistors/diodes. Just beware when you move to anything with Capacitors/Inductors, as impedance doesn't fit into the water-in-pipes picture.
  5. Apr 3, 2012 #4
    As an electrical engineer, I think understanding circuits is much more important than calculating the voltages and currents and there is beauty in analyzing electric circuits too. Of course circuits have their own rules, perhaps not known to some of physics students.

    As an example, suppose we have a network of 20 resistors and 5 voltage sources and suppose that we need to know the current in one of the resistors only. One can starts with KCL and KVL and form a large system of equations to find the current on the chosen resistor, however there is much better way to do so, and that's what we know as Thevenen's and Northon's equivalent circuits. In this method the rest of the circuit ( 5 voltage source and 19 resistors) is modeled with a single resistor and a single voltage source, the two in series with the chosen resistor. Constructing the equivalent circuits is much simpler than the KCL-KVL approach. Now its very easy to calculate the current in the resistor. More importantly, if the resistor changes, the equivalent of the rest of the circuit which we already built, wont change and calculating the current as a function of the resistor becomes very convenient.
  6. Apr 3, 2012 #5
    I guess part of the problem has to do with how quickly my professor covered all the circuit chapters.

    And I suspect the other problem is that I'm a visual learner. When I tackle the problem I usually close my eyes and imagine the equations and particles for a bit or draw a picture.

    Well, there is a lot of things I don't understand.

    1) How do these imaginary loops work. I'm talking of the ones you place arbitrarily in or around a circuit.
    2) Sign Conventions
    3) What is really going on within the circuits. For example, the resistors, emfs, potentials, loops, etc.
    4) Why is that when you travel from a - to +, the emf is considered to be positive? And vice-versa?

    Thanks, that helped a lot.

    Oh, so the circuits introduced in the book have a lot of passive components? I never thought about it that way really.

    Excuse my incompetence with circuits and resistors-- but what does it mean to go from a direction of high potential to low in a resistor? I thought resistors are just a part where things get slowed down a bit. What is going on within the resistor?
  7. Apr 3, 2012 #6
    I like understanding it electrically though.

    That is impressive.

    And no doubt that there is beauty in circuits. But I stand no chance of seeing it until I completely understand them-- until then, they will be a bit painful to deal with.
  8. Apr 3, 2012 #7
    If you get a chance and want to see some real beauty in circuits, look at the application of dispersion relations (Kramers kronig) to passive electric circuits in Vol II of Morse and Feshbach Methods of Theoretical Physics or in Bode'e book Network Analysis and Feedback Amplifier Design
    For example, Real Part Sufficiency means that if the real part impedance of a passive circuit is known at all frequencies, the imaginary part impedance is completely determined.
  9. Apr 3, 2012 #8
    Suppose we have a series RLC circuit. We know the real part R, at all frequencies ( it's a constant), how can we determine the imaginary part?

    Perhaps I didn't get you.
  10. Apr 3, 2012 #9


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    Then it would seem that circuits is a natural fit. Circuits are usually presented very visually.

    They work by Kirchoff's voltage law. That is one of the two fundamental laws of circuit analysis and is itself a limiting case of Maxwell's equations.

    Not all circuits can be analyzed using loop-current analysis. I generally prefer node-voltage analysis for that reason.

    The sign conventions are just that, conventions. There is no logical reason for any of them to be the way they are. If you consistently applied the opposite conventions you would get the correct answers, but don't do that or you will just confuse yourself and others.

    This is too broad. Can you narrow your question?

    Because the change from - to + is positive. I.e. if you have a - and you need a + then you have to add a +, not a -.
  11. Apr 3, 2012 #10
    The Kramers Kronig relations are discussed in Jackson Classical Electrodynamics (2nd Edition) page 311.

    See the Bode relations between real and imaginary components in

    http://ticsp.cs.tut.fi/images/2/23/Cr1023-riga.pdf Equations 7 and 8.

    See also for example the paper
    by Michael S. Wengrovitz', Alan V. Oppenheim', and George V. Frisk

    "A well-known property in Fourier transform theory is
    that causality in one domain implies real-part sufficiency
    in the alternate domain. This property is the basis for
    the fact that the real and imaginary components of a sig-
    nal are related via the Hilbert transform, if the spectrum
    of the signal is causal."

    See Eq (7) and (8).

    There are other sources, but they are pay per view.

    Your example, a series RLC circuit, is a very special case.

    A better example of a simple circuit is the parallel LR circuit, which has a complex impedance
    [tex] Z(\omega)=\frac{\omega^2 L^2 R+j\omega L R^2}{R^{2}+\omega^{2} L^2} [/tex]
  12. Apr 3, 2012 #11
    Thanks Bob S,

    This is a was very interesting property.
    As I gathered, one of the requirement for applying the rule is that both real and imaginary parts of the impedance must vanish faster than [itex]\frac{1}{|\omega |}[/itex] which is not satisfied in an RLC circuit.
  13. Apr 4, 2012 #12
    The Bode relations do apply to RLC circiuits, with some restrictions. As can be seen in Bode's relations

    http://ticsp.cs.tut.fi/images/2/23/Cr1023-riga.pdf Equations 7 and 8.
    There are constants of integration in the relations.

    The real part is positive definite, and includes ony even powers of ω; ω0 (constant), ω2, .......
    The imaginary part can be either ±j, and includes only odd powers of ω; ω-1, ω+1, ...............
    For a parallel LR circuit,
    [tex] Re[Z(\omega)]=\frac{\omega^2 L^2 R}{R^2 + \omega^2 L^2}[/tex]
    [tex] Im[Z(\omega)]=\frac{\omega L R^2}{R^2 + \omega^2 L^2} [/tex]
  14. Apr 4, 2012 #13


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    1) The loops are usually drawn to indicate the direction (or assumed direction) of the flow of charges (a.k.a. the current) in the circuit. Note that we use conventional current, which is regarded as being in the direction of the flow of positive charges, even though we know that in a wire, the charge carriers are actually negative charges. In the majority of situations, there is no difference between regarding it as a flow of positive charges in one direction or a flow of negative charges in the opposite direction.

    2) See my explanation of resistors below.

    3) I agree with DaleSpam. This question is too vague.

    4) You're moving from lower electric potential to higher electric potential. So your change in electric potential is positive. Or, to put it another way, the potential difference, or voltage, between the point labelled "+" and the point labelled "-" is positive.

    Passive components are things like resistors, inductors, capacitors, maybe even diodes, that don't require additional external power in order to work. Active components are electronic (usually semiconductor) devices like transistors, and everything that you can make using transistors, including operational amplifiers, voltage comparators, and innumerable other things. These things do require external power. Active electronics are not really covered in the circuit section of a physics course. You typically only learn about them in a dedicated electronics course in an electrical engineering program. I only really brought them up in order to make the point that if you actually wanted to build a current supply or a voltage supply that functioned somewhat like its "idealized" counterpart, then you would need to use active electronics.

    The purpose of introducing the "idealized" voltage source or current source is akin to the purpose of introducing the frictionless plane or the massless rope: by removing unnecessary complications, you can focus on learning basic principles. An example of the non-ideality: a simple battery doesn't quite function as an ideal constant voltage source, because it has some internal resistance, which means that its terminal voltage depends on the load current you draw from it. Its terminal voltage also declines with time as the battery discharges. To take into account the first problem, you can model a battery as an ideal voltage source in series with a resistor that represents the battery's internal resistance. So, as you can see, the idealized circuit devices can be useful for modelling the behaviour of real-world devices.

    KVL says that if you just have a voltage source in series with a resistor (and nothing else in the circuit), then any potential gained in going across the voltage source has to be lost in going across the resistor, so that the net change in potential going around the loop is zero. Therefore, there must be a difference in electric potential across the resistor. A potential difference is also known as a "voltage." This voltage across the resistor always goes down (drops) in the direction in which the current is flowing. In other words, the charges have more potential energy "upstream" of the resistor than "downstream" of it. Where does this energy go? Charges, in flowing across a resistor, lose energy to collisions with the nuclei of the atoms in the crystal lattice of the resistive material. Said nuclei enter into vibrational motion. So the energy goes into the small scale random motions of the individual particles in the material. What's another term for energy associated with that kind of random atomic/molecular motion? Thermal energy. In other words, the energy of the charges gets dissipated as heat in the resistor.

    The sentence in bold above is the rationale for the sign convention for resistors. Whatever direction you choose for the voltage drop across the resistor determines the direction of the current flow across it, or vice versa. I've attached a diagram showing this.


    EDIT: As you can see above, the potential at the top point where current enters the resistor is indicated as being higher than the potential at bottom point where the current leaves the resistor. The direction of the current is the same as the direction of the voltage drop. Suppose that V0 = +5 V and R = 1 Ω. Then KVL says VR = V0 = +5 V, and Ohm's law says that I = VR/R = +5 A, where a positive current means, "in the direction indicated by the arrow."

    So why do we keep telling you that it is just a convention? Aren't these directions the only right ones to choose in this situation, because of the direction of the EMF? Well, suppose I instead drew my resistor with the current and voltage drop indicated in the opposite direction, like so:

    Code (Text):

    \  -  ^
    /      |
    \      |  
    / +   |

    Then KVL would tell me that:

    V0 + VR = 0

    VR = -V0 = -5 V

    So the voltage at the bottom of the resistor is "higher" than the voltage at the top by -5 V. In other words, the voltage at the top is really higher than at the bottom by 5 V (which is the same answer as we got using the opposite assumed signs).

    Similarly, the current is I = VR/R = -5 V / 1 Ω = -5 A.

    So the answer for the current comes out negative, meaning that it flows in the direction opposite to what was indicated by the arrow. It flows from the top of the resistor to the bottom. This is exactly the same answer as what we got before.
    Last edited: Apr 4, 2012
  15. Apr 5, 2012 #14
    After some practice, I noticed that it really isn't bad at all. Its just some energy of conservation principles. Part of it is that I spoke too soon :X, and the other is because the professor went really quickly over the circuit chapters and I missed a lecture and failed to follow up at home.

    Anyways, thanks to all for the advice. :approve:
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