# Applying linearity in a circuit

• Engineering
• rms5643
In summary, the problem involves finding Io in the network shown in the figure using linearity and the assumption that Io = 1 mA. After calculating various values, including VR1, VR2, VR3, I1, I2, VR4, and VS, the focus is on finding I3. The attempt to solve for I3 involved using Ohm's Law and assuming that the voltage drop across the two resistors on the left summed to 208v. However, this resulted in a 3mA calculation error due to rounding. The correct calculation, using fractions instead of decimals, is 3.05mA, within the +/- 2% tolerance of the online homework. Overall, the methodology was verified as correct.
rms5643

## Homework Statement

Find Io in the network in the figure below using linearity and the assumption that Io = 1 mA.
Figure:http://i.imgur.com/Xtu0VmG.jpg

## Homework Equations

KCL, KVL, basic analysis techniques.

## The Attempt at a Solution

The following values I have calculated correctly:
VR1=9V
VR2=61V
VR3=70V
I1=1mA
I2=2mA
VR4=138V
VS=208v
I3=??
IS=??
IoActual=??

So, my trouble lies in finding I3, as once I find I3, I can easily find IS using KCL, though, the current source is doing something to the voltage drop across I3 that I'm not understanding.

How I tried to solve for I3 was the same way I solved for VS. Since the voltage drop across the source is 208v, then I assumed that the voltage drop across the two resistors on the left summed to 208v as well.

So, using Ohms Law: V/R=I, 208v/(42000Ω+26000Ω)=3mA, which is incorrect. Where did I go wrong?

Thank you

Rounding errors probably. Your analysis is fine. Just use fractions instead of calculating decimals and rounding.

1 person
vela said:
Rounding errors probably. Your analysis is fine. Just use fractions instead of calculating decimals and rounding.

Thank you very much Vela. You were correct. 3mA was incorrect, but 3.05mA was correct. (My online homework has a +/- 2% tolerance, so 0.5 was enough to fall outside of the threshold!

Thanks for verifying my methodology.

## 1. What is linearity in a circuit?

Linearity in a circuit refers to the property of a circuit where the output is directly proportional to the input. This means that if the input value is doubled, the output value will also double. In other words, the response of the circuit is a straight line when plotted on a graph.

## 2. How is linearity applied in a circuit?

Linearity is applied in a circuit by using linear components such as resistors, capacitors, and inductors. These components have a linear relationship between the voltage and current, which allows for a predictable and proportional response in the circuit.

## 3. What are the benefits of applying linearity in a circuit?

Applying linearity in a circuit allows for easier analysis and design, as the response of the circuit is predictable and follows a linear relationship. This also allows for easier troubleshooting and debugging of the circuit.

## 4. How does non-linearity affect a circuit?

Non-linearity in a circuit can cause distortion in the output signal, as the response is not proportional to the input. This can result in inaccurate measurements or unexpected behavior in the circuit. Non-linear components, such as diodes and transistors, can also introduce noise and introduce non-linearities in the circuit.

## 5. Can non-linear elements be used in a circuit that requires linearity?

Yes, non-linear elements can be used in a circuit that requires linearity, but they must be carefully chosen and configured. For example, diodes can be used to create non-linear responses in a circuit, but their behavior can be controlled by using feedback circuits or by using them in a specific configuration to achieve the desired linearity.

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