Saturday, 23 February 2013
SUPER NODE analysis
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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.
First mark a reference node v0. Then mark all other nodes
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
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1 Responses to “SUPER NODE analysis”
23 June 2018 at 01:40
What would happen if 10ohm resistor connected in series with 10 ohm resistor
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