# Archived Superposition to solve circuit with dependent source

1. Feb 24, 2014

### zealeth

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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: Feb 25, 2014
2. Feb 5, 2016

### Staff: Mentor

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:
$V_s - I R_1 = (I + I_s) R_2 + 2 I$

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