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Start for freeIntroduction to Network Equations
In circuit analysis, there are two primary methods for writing network equations:
- Mesh analysis
- Node analysis
Both methods allow us to solve for unknown currents and voltages in a circuit. The choice between mesh and node analysis often depends on which method requires fewer equations for a given circuit.
Mesh Analysis
Mesh analysis involves identifying closed paths in a circuit called meshes and writing equations based on Kirchhoff's Voltage Law (KVL) for each mesh.
Key Points:
- A mesh is the smallest closed path that does not contain any other closed path within it
- All meshes are loops, but not all loops are meshes
- The number of independent mesh equations required is: B - N + 1 Where B = number of branches, N = number of nodes
- Each circuit element must be included in at least one mesh equation
Example of Mesh Analysis
Let's consider a circuit with two meshes:
+ L C1 R1
e(t) --- ))) ---||--- /\/\/\ ---+
- |
C2 |
---||--- |
L2 |
))) |
|
R2
/\/\/\
|
---
GND
For this circuit, we would write two mesh equations:
-
For mesh 1 (i1): e(t)u(t) = L(di1/dt) + (1/C1)∫i1dt + R1(i1-i2) + (1/C2)∫(i1-i2)dt
-
For mesh 2 (i2): 0 = L2(di2/dt) + R2i2 + (1/C2)∫(i2-i1)dt + R1(i2-i1)
Node Analysis
Node analysis involves writing equations based on Kirchhoff's Current Law (KCL) for each node in the circuit, except for the reference node (usually ground).
Key Points:
- The number of independent node equations required is: N - 1 Where N = number of nodes
- Node voltages are measured with respect to a reference node
Example of Node Analysis
Consider a circuit with three nodes:
i(t)u(t)
|
R0
|
+---+---+
| |
C0 R1
| |
| L1
| |
+---+---+
|
C1
|
R2
|
GND
For this circuit, we would write three node equations:
-
For node 1 (V1): i(t)u(t) = V1/R0 + C0(dV1/dt) + (V1-V2)/R1 + (1/L)∫(V1-V3)dt
-
For node 2 (V2): 0 = (V2-V1)/R1 + C1(dV2/dt) + V2/R2
-
For node 3 (V3): 0 = (1/L)∫(V3-V1)dt + C1(d(V3-V2)/dt)
Initial Conditions
When solving differential equations derived from circuit analysis, we need to consider initial conditions. These are typically evaluated at t = 0+, just after a switch is closed or a source is applied.
Methods for Finding Initial Conditions:
- Physical considerations
- From the differential equations themselves
Example: RC Circuit
For a simple RC circuit with a voltage source V switched on at t=0:
+ R
V --- /\/\/\ ---+
- |
C
|
---
GND
- i(0+) = V/R (capacitor acts as a short circuit initially)
- di/dt(0+) = -V/(R^2C) (found by differentiating the circuit equation)
Steady-State Solutions
The steady-state solution is the long-term behavior of the circuit as t approaches infinity. For DC sources, this often involves treating inductors as short circuits and capacitors as open circuits.
Sinusoidal Steady-State
For sinusoidal sources, we can use phasor analysis to find the steady-state solution without solving differential equations.
Example: Current source parallel with RC
i0sin(ω0t)u(t)
|
+---+---+
| |
R C
| |
+---+---+
|
GND
The voltage across the parallel RC combination can be found using phasor analysis:
V(t) = (i0 / √(G^2 + (ω0C)^2)) * sin(ω0t - tan^(-1)(ω0C/G))
Where G = 1/R
Conclusion
Understanding how to write and solve network equations using mesh and node analysis is crucial for circuit analysis. By considering initial conditions and steady-state solutions, engineers can fully describe the behavior of a circuit over time. For sinusoidal sources, phasor analysis provides a powerful tool for finding steady-state solutions without solving complex differential equations.
Article created from: https://youtu.be/gk7HNFBXi_c?si=ONobVY576AjK6lyP