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This is a lab report made using MATLAB circuits and its results to prove and work on Thevenin Circuits.
Typology: Slides
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→ Measure open circuit voltage at CG by removing the 150 Ω resistor in the network given above. → Short circuit the existing voltage source then apply 1 voltage source across CG to find the RTh.
Results ➢ Current through RL=0.006583 A (in actual circuit)
Inference
Thevenin’s Theorem is basically used to calculate current through a load whose resistance value keep changing. We convert a (difficult to analyse again and again) linear two terminal circuit into a simple circuit with single voltage source (Vth)and a resistance (Rth) in series. → In the circuit given, we obtain Vth (Thevenin's voltage) =1.945V → We obtain Rth (Thevenin’s resistance) =145.5 ohms by short circuiting the voltage source. → And by using Vth and Rth we can calculate current through RL(6.582mA) which is approximately equal to current through RL(6.583mA) in actual circuit. Here Vth is open circuit voltage(Va-b) and Rth is equivalent resistance at the terminals (a and b) when all the independent sources are turned off. Conclusion In this experiment we learnt about the Thevenin Theorem and also verified it for the given circuit. Also, we learnt the application of this theorem and how this method is useful to calculate current at points where it is difficult to calculate the current by usual methods. ▪ Vth = 1.9453 V ▪ Rth = 145.5 Ω
Procedure → Connect the circuit as shown in the figure 3 and 4 and then measure the current i 1 and i 2 respectively. → Find out the value of current i in the circuit shown in figure 2. → Verify the superposition theorem from the observations. MATLAB Circuit
Results → i=0.0029924 A
For verification of Superposition Theorem, i= i 1 + i 2
According to Superposition Theorem, in linear circuits (linear sources and linear elements) the response at any point in a linear circuit having more than one independent source is equal to sum of the responses when each source acting alone. → In the circuit given, current through AB branch when only V 1 is active is 12.568 mA while current through AB when only V 2 is active and V1 is sorted is – 9.5756 mA. → Individual contribution by both the sources when added give us same result as the actual circuit when both the sources are active, that is 2.9924 mA.
In this experiment, we learnt about the superposition principle and verified it. This method can be used to find voltage and current at any point in a very simple way in any linear circuit. ▪ i 1 = 0.012568 A ▪ i 2 = - 0.0095756 A ▪ i = 0.0029924 A
