Doubt in Superposition Theorem

Discussion in 'General Electronics Chat' started by logearav, Jan 1, 2013.

  1. logearav

    Thread Starter Member

    Aug 19, 2011
    248
    0
    Dear Members,
    I have this doubt in superposition theorem discussed threadbare in AAC material. Kindly go through my attachments. I can't understand how those E values for R1, R2 and R3 were arrived at? If i know this i can sail thro the topic which is well written.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    But what this simply circuit has to do with superposition theorem?
    Simply use Ohms law and two Kirchhoff's laws to solve this circuit.
     
  3. logearav

    Thread Starter Member

    Aug 19, 2011
    248
    0
    http://www.allaboutcircuits.com/vol_1/chpt_10/7.html
    Thanks for the reply. My attachment was taken from this topic. Though Kirchoff laws are useful, this circuit is described to explain the superposition theorem. I can't understand this first step which i given in my original post. If explanation is provided i can proceed further.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    So you don't understand why they remove from the circuit the 7V battery?
    Or you simply don't now how to solve this circuit (that with one 28V battery)?
     
  5. mwalden824

    Member

    Mar 6, 2011
    51
    2
    The first step you posted was done by short circuiting the right side voltage supply. This is done for all independent supplies but if it is a independent current supply, then you replace it with an open circuit. The effect on the circuit from each supply alone is determined and then the effects are added together to get the total effect.

    So first you would short one of the supplies, then solve the EIR chart, then short the other supply and repeat the process again, but obtaining different values of course. Then you take these two charts and add them together but noting the polarities of the currents and voltages, because if the current flows through a resistor in the same direction for both of the independent voltage supplies, then they add, but if they flow in opposite direction they subtract.

    Later,
    Michael
     
    logearav likes this.
  6. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    I think the problem comes down to looking at how to use the principle of superposition without understanding what it is all about.

    For a linear circuit (and superposition only applies to linear circuits), the solution for every voltage or current in the circuit will be of the form:

    <br />
x = AV_1 + BV_2 + CI_1 + DI_2<br />

    for a circuit with two independent voltage supplies and two independent current supplies.

    If x is a voltage, the A and B will be dimensionless and C and D will have units of resistance. If x is a current, then A and B will have dimensions of conductance while C and D will be dimensionless. The important this is that the quantity of interest, x, will be a linear sum of terms with each term involving exactly one of the independent supplies.

    Now, this can be written in terms of four partial results

    <br />
x = x_1 + x_2 + x_3 + x_4<br />

    where

    <br />
x_1 = AV_1<br />
x_2 = BV_2<br />
x_3 = CI_1<br />
x_4 = DI_2<br />

    If we can find the four simpler quantities, then we can just add them together to get the final result.

    But what does x_1 mean, physically? Well, what did the original expression for x mean, physically? Simple. It told us how to find the value of x as a function of different values for the outputs of the four independent sources. So what if we just choose to set all but one of these to zero? Again, simple. All we are left with is the single term that we did not set to zero.

    From a circuit standpoint, this is the same as setting the voltage on a voltage source to zero, meaning that any amount of current can flow through the supply but no voltage can appear across it. Well, that is the description of a piece of wire, so we can effectively set the voltage on a voltage source to zero by replacing it with a short circuit. Similarly, setting the current on a current source to zero means it can have any voltage across it, but can have no current through it, which is the description of an open circuit. So we turn off current supplies by replacing them with an open circuit.

    Thus, in its simplest form, we can turn off all the independent supplies and then turn one of them on and analyze all the quanties in the circuit we are interested in. Then turn that one back off and turn on another one and do the same thing. We keep doing this until each supply has been turned on exactly once and then add all the results together.
     
    logearav likes this.
  7. logearav

    Thread Starter Member

    Aug 19, 2011
    248
    0
    Thanks everyone for your replies. I don't understand how the values of EIR chart arrived at? For example when 7 v battery is shorted, the total effective resistance (R1,R2 and R3) is 14/3 ohm which is 4.666 ohm and that is given in the last row(R) of the chart. I = V/R so 28/4.666 which gives 6 Ampere and that is the first entry in the second row(I). Then how 2 Ampere comes in the second entry of (I) that is current value corresponding to R2 that is 4 ohm. The formula is I = V/R but what is the values for the V and R in this instance?
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
  9. Arventh

    New Member

    Feb 12, 2013
    1
    0
    What is the difference between a voltage source and a current source ?
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    A constant voltage source maintains a constant voltage between two nodes but lets any amount of current (source or sink) to flow through the source in order to achieve that.

    A constant current source maintains a concetant current between two nodes but lets any amount of voltage (source or sink) to appear across the source in order to achieve that.
     
Loading...