# Superposition to solve circuit with dependent source

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• zealeth
In summary, using the superposition principle and applying KCL and KVL, the current I in the circuit was found to be 2.024 Amps when the current source was suppressed and 0.357 Amps when the voltage source was suppressed. Combining these results, the final current was determined to be 2.024 Amps. The solution also involved using Ohm's Law and current and voltage dividers.
zealeth

## Homework Statement

Assume that Vs = 10V , R1 = 1.7Ω , and R2 = 0.50Ω. Find the current I in the figure using the superposition principle.

## Homework Equations

Superposition principle
KCL, KVL
V=IR
Current and voltage dividers

## The Attempt at a Solution

I = I' + I''

First replacing the current source with an open circuit, I have:

(1.7+0.50)*I' = 10 - 2I' using KVL
Solving, I' = 2.38 A

Now replacing the voltage source with a short, I use the current divider to get:

-I'' = 3*0.5/(1.7+0.5) = 0.682 A
I'' = -0.682 A

I = 2.38 - 0.682 = 1.7 A, which is incorrect.

EDIT: Should I have left the dependent source as 2I and solved for I in the final equation, or am I doing something else wrong?

EDIT2: Yes, I did have to leave it as 2I. Got it :)

Last edited:
A complete solution to wrap things up.

Using superposition:
1) Suppressing the current source:

We are left with a single series circuit (blue dotted components are suppressed/removed):

Writing Ohm's Law for the circuit we have
## I' = \frac{V_s - 2 I'}{R_1 + R_2} ##

## I' = \frac{V_s}{R1 + R2 + 2}~~ ## after solving for I'

## I' = 2.381~Amps##

2) Suppressing the voltage source leaves us with:

Let v be the voltage at the junction of the resistors. Then writing a node equation there:

##\frac{v}{R_1} - I_s + \frac{v - 2 I''}{R_2} = 0##

But Ohm's Law also gives us ##v = -I'' R_1##. So substituting for v in the node equation:

## -I - I_s + \frac{(-I'' R_1 - 2 I'')}{R_2} = 0##

##I'' = -I_s \frac{R_2}{R_1 + R_2 + 2} ##

##I'' = -0.357~Amps##

3) Combining the results:
##I = I' + I'' = 2.024~Amps##

Quick Check:
##V_s - I R_1 = (I + I_s) R_2 + 2 I##

##I = \frac{V_s - I_s R_2}{R1 + R_2 + 2} = 2.024~ Amps##

#### Attachments

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Greg Bernhardt

## 1. How is superposition used to solve circuits with dependent sources?

Superposition is a problem-solving technique that involves breaking down a circuit into smaller, simpler sub-circuits and analyzing each one separately. This method is especially useful when dealing with circuits that have dependent sources, as it allows us to determine the effect of each source on the circuit's overall behavior.

## 2. What is a dependent source in a circuit?

A dependent source is a type of circuit element whose output is determined by some other parameter or component in the circuit. This can be a voltage or current source that is dependent on the voltage or current of another source or element in the circuit.

## 3. What is the first step in using superposition to solve a circuit with dependent sources?

The first step is to identify all the sources in the circuit, both independent and dependent. Then, we need to turn off all but one of the independent sources and analyze the resulting circuit. This will give us the contribution of that source to the overall circuit behavior.

## 4. Are there any limitations to using superposition to solve circuits with dependent sources?

Yes, there are some limitations to using superposition. This method can only be applied to linear circuits, meaning that all the circuit elements must follow Ohm's law and have a linear relationship between voltage and current. Additionally, it can only be used for circuits with a single frequency source.

## 5. Can superposition be used for circuits with multiple dependent sources?

Yes, superposition can be used for circuits with multiple dependent sources. In this case, we would need to repeat the process for each dependent source, turning off all the other sources and analyzing the circuit separately. Then, we can combine the results to determine the overall behavior of the circuit.

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