Help with superposition circuit analysis!?

Question

http://i143.photobucket.com/albums/r147/rAOL87/Circuit3.jpg

The circuit above is the circuit I must analyze. I've been told that the current that runs through the points Va and Vb is zero because of the symmetry. I need to find the individual currents and voltages that run through each resistor, and I was wondering how I go about this with superposition?

Answer 1

Voltage at nodes a & b is 2.5 V -for both- from inspection. It's true, the current in the resistor spanning a, b, is zero, since both nodes are at the same potential. You can as well remove this resistor from the circuit; operationally, it'll make no difference. However, if you do, it should be clear enough that all that remains is a pair of equal, simple voltage dividers.

At this point, it should became clear, as well, that the voltage at nodes a, b, is just half the source voltage, namely 2.5 V. From symmetry considerations, current in every (remaining) resistor is the same, 2.5 / 3.3K = 757.6 µA.

About the second part of your question, suppose you want to verify the voltage in nodes a, b, is just 2.5 V, by superposition. Since the circuit has two sources, this calculation will involve as many stages. Superposition implies reckoning the effect each source has upon the circuit, one at a time.

Accordingly, "kill" the rightmost source, replacing it by a short circuit. Consider node a. Let's find out which its voltage should be. Proceeding right-to-left, we find two 3.3K resistors in parallel. This combination has an equivalent resistance of 1.65K. Next, there's another 3.3K resistor, in series with the parallel set. The eq. resistance, 4.95K.

The latter is shunted by another 3.3K resistor. Eq. resistance so far: 1.98K. Finally, we find another resistor in series. Total eq. resistance: 5.28K. The current supplied by the 5 V source: 947 µA. Va is just 5 V - 3.3K × 947 µA = 1.875 V.

This is the share the left source contributes to total voltage at node "a". To this, you should add the share the other source puts out. In order to do this, "kill" the right source, then proceed as above, this time left-to-right. You should find that the voltage at node "a", due solely to the right source, is 0.625 V. You can now "superpose" the effect of both sources, yielding 2.5, Q.E.D.

Since the circuit is completely symmetrical, voltage at node "b" is just the same as for node "a".

Answer 2

Superposition Circuit Analysis

Answer 3

Yeah ,
Since The current in path AB is o because of symmetry.
Using Superposition principle.
Thus the wire below AB is same as grounded.

Consider the left hand 5V cell,
let current i flow through this circuit.
Net resistance=3.3+3.3=6.6k ohms
Current=V/R=5/6600=7.575 * 10^-4 A
Since resistances are in series current through both of them is 0.7575 mA

Similarly for the right hand half of the circuit
Answer is the same = 0.7575 mA

Answer 4

Thevenin idea is you may eliminate the source. once you eliminate a voltage source, the voltage will develop into 0. the in effortless words thanks to ascertain that the voltage drop between to factors is 0 is to short them mutually once you eliminate a modern source, the present will develop into 0.. the in effortless words thanks to ascertain that the present in a route is 0 is to introduce an open circuit into that route.