Thevenin and Norton Theories help

In summary, the problem presents a wheatstone bridge design with a 10 V DC power supply, where R1 = 2 ohms, R2 = 3 ohms, Rx (presented as R4) = 4 ohms, and the typical 'bridge' portion is replaced with a 1 ohm resistor (labeled R3). The load is placed at the location of R3 and has a resistance of 5 ohms. The task is to find the Thevenin model for this circuit, including Vth and Rth. The instructor's answers are Rth = 1.42 ohms and Vth = 6.128 V. However, the student's calculations resulted in Rth =
  • #1
classyfella
1
0
the problem presents a wheatstone bridge design, with a 10 V DC power supply. R1 is 2 ohms, R2 is 3, Rx (presented as R4 for this problem) is 4, the typical 'bridge' portion is replaced with a 1 ohm resistor (labeled R3), and where you usually have R3 you have a load, which is 5 ohms.
The problem asks to present a thevenin model for this same probelm, Vth and Rth.
My instructor has his answers posted, but not how he got them.

My answers came out with Rth = R1||R2+R3||R4 = 2||3+4||1 = 2
my Voltage obtained came out to Vth = R!/(R1 + R2||(R3+R4)) * Vs = 2/(2+3||(1+4))*10=5.16V
however for Rth my instructor has 1.42 ohms and for Vth he has 6.128

Can anyone help me see what I did wrong, how to get the values my teacher got (as I haven't been able to figure it out.) and possibly give any tips/tricks for thevenin and norton, as these guys are giving me a lot more trouble than I care for.
 
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  • #2
I'd need to see the schematic.
 
  • #3
You should post the pic of your question kinda becomes hard to visualize (rather lazy):)
 

1. What is the Thevenin and Norton theorem?

The Thevenin and Norton theorem is a fundamental theory in electrical engineering that states that any linear, active bilateral network can be represented by an equivalent circuit consisting of a voltage source and a series resistance (Thevenin equivalent) or a current source and a parallel resistance (Norton equivalent).

2. What is the purpose of using Thevenin and Norton equivalent circuits?

The Thevenin and Norton equivalent circuits allow for simplification and analysis of complex networks, making it easier to determine the behavior of a circuit and its components. It also allows for the replacement of a portion of a circuit with a simpler, equivalent circuit, without changing the overall behavior of the original circuit.

3. How do I calculate the Thevenin and Norton equivalent values?

To calculate the Thevenin equivalent, you need to find the open-circuit voltage and the equivalent resistance of the original circuit. The open-circuit voltage is the voltage across the load terminals with the load disconnected, and the equivalent resistance is the resistance seen from the load terminals with all sources turned off. To calculate the Norton equivalent, you need to find the short-circuit current and the equivalent resistance. The short-circuit current is the current through the load terminals with the load connected by a short circuit, and the equivalent resistance is the resistance seen from the load terminals with all sources turned off.

4. Are there any limitations to using Thevenin and Norton equivalent circuits?

The Thevenin and Norton equivalent circuits are only valid for linear, active, and bilateral networks. They also assume that the circuit parameters do not change with time, temperature, or frequency. Additionally, the equivalent circuit may not accurately represent the behavior of the original circuit if the load is significantly different from the original load.

5. How are Thevenin and Norton equivalent circuits used in practical applications?

The Thevenin and Norton equivalent circuits are frequently used in the analysis and design of electronic circuits, such as amplifiers and filters. They are also used in power systems to determine the maximum power transfer and to model the behavior of transmission lines. Additionally, they are used in simulation software to model and simulate complex circuits.

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