Saturday 23 February 2013

SUPER NODE analysis


When a voltage source comes in between two  node then these two nodes and the voltage source form a supernode and we take this supernode as a single node and apply KCL and KVL .
Rules for solving a supernode
1 Mark a reference node such that a supernode can’t be formed. Try to avoid supernode at first hand. If it’s not possible then at least make a voltage source referenced.

2) Mark other non-referenced nodes as you do in normal nodal analysis.

3) Mark the supernode with a dotted circle to remind you that it’s a super node.
4) Now apply KCL at the super node.
5) Apply KVL at the super node loop to find the node voltages in super node.

Example:


First mark a reference node v0. Then mark all other nodes
Now we have marked a dotted circle to denote a super node along with V1 and V2.


Remember that the 10 Ohm resistor connected across the supernode does not have any significance in the calculations as it is connected across the supernode
Now apply KCL to the circuit:
 (V1 – 0) / 2 + (V2 – 0) / 4 +7 = 2
2 V1 + V2 +28 = 8
2 V1 + V2 = -20 ………………………………………………. 1
Now apply KVL to the super node loop:
-V1 - 2 +V2 = 0
V2 = V1 + 2 ……………………………………………………. 2
Put equation 2 in equation 1
2 V1 + (V1+ 2) = -20
3 V1 = -22
V1 = -22/3 = -7.33 v
Now from equation ( 2 ) we get V2 = - (22 / 3) +2 = -16 / 3 =  -5.33 v.
                            V1  = -7.33 v
                            V2  = -5.33 v


1 Responses to “SUPER NODE analysis”

Unknown said...
23 June 2018 at 01:40

What would happen if 10ohm resistor connected in series with 10 ohm resistor


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