THIS BLOG CONTAIN ABOUT THE THEVENIN THEOREM IN AC CIRCUIT.
WHAT IS THEVENIN THEOREM IN AC CIRCUIT?
Thévenin's Theorem for AC circuits with sinusoidal sources is very similar to the theorem we have learned for DC circuits. The only difference is that we must consider impedanceinstead of resistance. Concisely stated, Thévenin's Theorem for AC circuits says:
Any two terminal linear circuit can be replaced by an equivalent circuit consisting of a voltage source (VTh) and a series impedance (ZTh).
In other words, Thévenin's Theorem allows one to replace a complicated circuit with a simple equivalent circuit containing only a voltage source and a series connected impedance. The theorem is very important from both theoretical and practical viewpoints.
It is important to note that the Thévenin equivalent circuit provides equivalence at the terminals only. Obviously, the internal structure of the original circuit and the Thévenin equivalent may be quite different. And for AC circuits, where impedance is frequency dependent, the equivalence is valid at one frequency only.
Using Thévenin's Theorem is especially advantageous when:
· we want to concentrate on a specific portion of a circuit. The rest of the circuit can be replaced by a simple Thévenin equivalent.
· we have to study the circuit with different load values at the terminals. Using the Thévenin equivalent we can avoid having to analyze the complex original circuit each time.
We can calculate the Thévenin equivalent circuit in two steps:
1. Calculate ZTh. Set all sources to zero (replace voltage sources by short circuits and current sources by open circuits) and then find the total impedance between the two terminals.
2. Calculate VTh. Find the open circuit voltage between the terminals.
EXAMPLE
In this circuit, the load is the series-connected RL and CL. These load components are not part of the circuit whose equivalent we are seeking. Find the current in the load using the Norton equivalent of the circuit.
v1(t) = 10 cos wt V; v2(t) = 20 cos (wt+30°) V; v3(t) = 30 cos (wt+70°) V;
v4(t) = 15 cos (wt+45°) V; v5(t) = 25 cos (wt+50°) V; f = 1 kHz.
First find the open circuit equivalent impedance Zeq by hand (without the load).
Numerically
ZN = Zeq = (13.93 - j5.85) ohm.
Below we see TINA's solution. Note that we replaced all the voltage sources with short circuits before we used the meter.
Now the short-circuit current:
The calculation of the short-circuit current is quite complicated. Hint: this would be a good time to use Superposition. An approach would be to find the load current (in rectangular form) for each voltage source taken one at a time. Then sum the five partial results to get the total.
We will just use the value provided by TINA:
iN(t) = 2.77 cos (w×t-118.27°) A
Putting it all together (replacing the network with its Norton equivalent, reconnecting the load components to the output, and inserting an ammeter in the load), we have the solution for the load current that we sought:
By hand calculation, we could find the load current using current division:
Finally
I = (- 0.544 - j 1.41) A
and the time function
i(t) = 1.51 cos (w×t - 111.1°) A
LEARNING
- In solving thevenin theorem first we find the ZTH and VTH.
- In finding ZTH we need to turn off all the sourcing.
- We can use the series and parallel in solving the ZTH.
- In solvong the VTH we can use the mesh and nodal analysis.
- We can use the basic law in solving the VTH.
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