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Lecture Nine
Nodal Analysis
Independent (Current) Sources
Problem solving strategy:
- Reduce the circuit to minimum number of nodes.
- The variables in the circuit are selected to be node voltages. These voltages are defined w.r.t. a common point in the circuit.
- One node is selected as reference node. Usually this node is one to which maximum number of branches are connected. It is commonly called ground and said to be at zero‐ potential.
- Select the variables as positive w.r.t. ground (reference node)
- Set the direction of all currents toward ground except that shown by current sources and write equations.
- Solve the equations for unknown using (i) Guass Elimination Method (ii) Matrix Analysis. Make sure that number of linearly independent equations matches the number of unknowns.
Example
Given that I (^) A = 1 mA, I (^) B = 4 mA, R 1 = 12 k Ω , R 2 = R 3 = 6 k Ω, calculate the node voltages and branch currents
The left‐ and the right‐most can be ignored in the analysis. Mark the nodes as 1 and 2 and mark the voltages w.r.t. ground as V 1 and V 2 respectively.
At node 1, due to direction, the current I (^) A is entering. Applying KCL at node 1
1 2 1 1 2 1 1 2 2 1 2 1 2 2 (^3 1 ) 1 2
4k 6k
I A I I
V V V V R R V
R R R R R
− V^ V^ V V
= −^ + −^ = + −
× = − ⇒ − =
At node 2
2 3 1 2 2 1 2 2 3 2 3 2 2 3 (^3 1 21 )
6k 3k
I B I I
V V V V V R R
R R R R R
− V^ V^ V V
= −^ − − = − +
× = − ⇒ − =
Now we get two set of equations namely
1 2 1 2
V V
V V
Subtracting them yields
1 2 1 2 1 1
6V
V V
V V
V
V
Similarly V 2 = ‐ 15 V. Now we can calculate current as
1 1 2 1 2 3 2 1 2 3
(^6) 0.5 mA; 6 15 1.5 mA; 15 2.5 mA 12k 6k 6k
I V^ I V^ V^ I V
R R R
= = −^ = − = − = −^ +^ = = = − = −
The final circuit becomes
Example:
Given R 1 = R 2 = 2 k Ω , R 3 = R 4 = 4 k Ω , R 5 = 1 k Ω , I (^) A = 4 mA, I (^) B = 2 mA. Calculate the node voltages.
Mark the nodes as 1, 2 and 3 and mark the voltages w.r.t. ground as V 1 , V 2 and V 3 respectively.
Note the direction of current source I (^) A that is away from ground. At node 1, due to current source I (^) A , the current I 3 (flowing through R 3 ) is leaving. Applying KCL, we get
3 5 3 5 3 1 3 2 1 2 3 3 5 3 5 3 5 3 5 (^3 1 231 2 )
4k 1k 4k
B B
I I I
I I I
V V V V V V V R R
R R R R R R
− V^ V^ V V V V
= − −^ − −^ = + − +
× = + − ⇒ + − =
Now we get 3 sets of equations
1 2 3 1 2 3 1 2 3
V V V
V V V
V V V
Using matrix analysis
1 2 3
0.7619 V
V
V
V
