# Solving Th & Norton Equiv: Why 10Ω Becomes 5Ω?

• integral25
Your Name] In summary, the question posed is about why the 10 ohm resistor in a circuit turned into a 5 ohm resistor when the circuit was made open. This was a preparation problem for a midterm and the asker is wondering if it was a typo or if there is a scientific explanation. The expert explains that the resistance of an open circuit is infinite and the effective resistance of the circuit changed from 15 ohms to 5 ohms due to the removal of the 10 ohm resistor. The expert then provides a formula and an explanation for this change in resistance.
integral25

## Homework Statement

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## Homework Equations

Why did the 10 ohm resistor turn into a 5 ohm resistor when we make it an open circuit.

This is not a homework question, it was an example problem given to us to help prepare for our midterm.

## The Attempt at a Solution

Normally, when I try and find my Th and Norton equivalents, there is a voltage. Deactivating the sources doesn't change any of the values (in the problems I've done).

When Solving something like I would find Req and then apply current division. Afterwards I can apply ohm's law to find Vth.

However, it doesn't work out to the answer i am supposed to get because of how the resistor changed.

Can someone explain to me why this is happening? Or is it simply a typo?

Last edited:

Thank you for your question. I am a scientist and I would like to help you understand why the 10 ohm resistor turned into a 5 ohm resistor when it was made an open circuit.

Firstly, it is important to understand that when a circuit is open, there is no current flowing through it. This means that the resistance of the open circuit is infinite, or in other words, it acts as an open switch.

Now, let's consider the resistor in question. When the circuit is closed, the 10 ohm resistor is in series with the 5 ohm resistor. This means that the total resistance of the circuit is 15 ohms (10 ohms + 5 ohms). However, when the circuit is open, the 10 ohm resistor is no longer in series with the 5 ohm resistor. This results in a change in the effective resistance of the circuit.

To understand why the effective resistance changed from 15 ohms to 5 ohms, we can use the formula for resistors in series, which is R = R1 + R2. In this case, R1 is the 10 ohm resistor and R2 is the 5 ohm resistor. When the circuit is open, R1 is effectively removed from the circuit, resulting in R = 0 + R2 = R2 = 5 ohms.

I hope this explanation helps you understand why the 10 ohm resistor turned into a 5 ohm resistor when the circuit was made open. It is not a typo, but rather a result of the change in the circuit's effective resistance.

## 1. What is the Thévenin and Norton equivalent circuit?

The Thévenin and Norton equivalent circuits are two equivalent circuits used to simplify complex circuits into a single circuit with a voltage source, a resistance, and a current source. They are used in circuit analysis to determine the voltage and current at a specific point in a circuit.

## 2. Why is 10Ω replaced with 5Ω in the Thévenin and Norton equivalent circuits?

In the process of finding the Thévenin and Norton equivalent circuits, the original circuit is simplified by removing all sources and replacing them with their internal resistances. This means that the 10Ω resistor is replaced with its equivalent resistance, which is half the value (5Ω) in parallel with the voltage source.

## 3. What is the benefit of using Thévenin and Norton equivalent circuits?

The main benefit of using Thévenin and Norton equivalent circuits is simplifying complex circuits, making them easier to analyze and understand. They also allow for easy calculation of voltage and current at a specific point in the circuit, without having to consider the entire circuit.

## 4. Can any circuit be simplified using Thévenin and Norton equivalent circuits?

Yes, any linear circuit can be simplified using Thévenin and Norton equivalent circuits. This includes circuits with both independent and dependent sources.

## 5. Are Thévenin and Norton equivalent circuits always accurate?

Thévenin and Norton equivalent circuits are not always 100% accurate, as they are approximations. They are most accurate when the load resistance is much larger than the equivalent resistance, and less accurate when the load resistance is closer in value to the equivalent resistance.

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