Superposition method – Circuit with two sources

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Electrical Circuits, Electrical Circuits Problems, Resistive Circuits

Find $I_x$ using superposition rule:

Solution

Superposition

The superposition theorem states that the response (voltage or current) in any branch of a linear circuit which has more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, while all other independent sources are turned off (made zero).

There are two sources, so we need to turn them off one by one and calculate associated $I_x$ values.

Two announcements before we start! First, a PDF sheet of this problem with the solution and side space for notes can be downloaded below. Second, you may watch solving this problem on the video posted below. My suggestion is that you print the solution sheet and make notes on it while watching the video. Please subscribe to my YouTube channel for more videos!

Let’s start with the current source. To turn a current source off we need to replace it with an open circuit. Easiest way to remember this is that to turn off a source we must make its value zero and for a current source we need to make its current zero. Zero current means no way for current to pass and that is possible with an open circuit.

Or simply

$4 \Omega$ and $2 \Omega$ are in series and also $3 \Omega$ and $1 \Omega$ :

Now $4 \Omega$ and $6 \Omega$ are parallel:
$4 \Omega || 6 \Omega = \frac{4 \times 6 }{4+6}=2.4 \Omega$

So using Ohm’s law:
$I_{x_1}=\frac{5 V}{2.4 \Omega}=2.083 A$

To continue solving the circuit with the Superposition method, we should make the voltage source zero and find the contribution of the current source on $I_{x}$ . Making voltage source means replacing it with a short circuit. Similar to the current source, you may memorize this by remembering that you need to make the source value (here: voltage) equal to zero and to force zero voltage drop between two points you need to connect them.

For a moment forget $I_{x}$ and concentrate on finding current of resistors. If we have the current of resistors, we can easily apply KCL and find $I_{x_2}$ . So, $4 \Omega$ and $2 \Omega$ are parallel and also $3 \Omega$ and $1 \Omega$ are parallel:

$4 \Omega || 2 \Omega = \frac{4 \times 2 }{4+2}=\frac{4}{3} \Omega$
$3 \Omega || 1 \Omega = \frac{3 \times 1 }{3+1}=\frac{3}{4} \Omega$

Now, we can find their voltage drops:
$V_{4 \Omega || 2 \Omega}=\frac{4}{3} \times -3 A= -4 V$
$V_{3 \Omega || 1 \Omega}=\frac{3}{4} \times -3 A= -2.25 V$
Please note that the voltage drop on $4 \Omega || 2 \Omega$ is the same as $4 \Omega$ and $2 \Omega$ voltage drops, because the circuits are equivalent and all are connected to the same nodes. The same statement is correct for $3 \Omega || 1 \Omega$ voltage drop and $3 \Omega$ and $1 \Omega$ voltage drops. So
$V_{4 \Omega}=V_{2 \Omega}=V_{4 \Omega || 2 \Omega}= -4 V$
$V_{3 \Omega}=V_{1 \Omega}=V_{3 \Omega || 1 \Omega}= -2.25 V$
To find $I_{x_2}$ all we need is to write KCL at one of the nodes:

$-I_{2 \Omega}+I_{x_2}+I_{3 \Omega}=0$
$\rightarrow I_{x_2}=I_{2 \Omega}-I_{3 \Omega}$

$I_{2 \Omega}$ and $I_{3 \Omega}$ can be found using Ohm’s law:
$I_{2 \Omega}=\frac{V_{2 \Omega}}{2 \Omega}=\frac{-4}{2}=-2A$
$I_{3 \Omega}=\frac{V_{3 \Omega}}{3 \Omega}=\frac{-2.25}{3}=-0.75A$
Therefore,
$I_{x_2}=-1.25 A$
And
$I_{x}=I_{x_1}+I_{x_2}=2.083-1.25= 0.8333 A$
Now, replace $1 \Omega$ resistor with a $6 \Omega$ one and solve the circuit using superposition method. Let me now your answer below in the comments section.

Solution sheet

Superposition method – Circuit with two sources – Solved Problems Download

KCL Superposition Superposition Method Superposition_Method Turning sources off

Comments

26 responses to “Superposition method – Circuit with two sources”

November 12, 2013

abjhns

0.33 A

Reply
1. January 4, 2015
  
  kashif ahmad
  
  how
  
  Reply
June 7, 2014

mahmoud SAAD

but if we use the mesh analysis we will get -2.08 A …
and ix2=-1.25 A
SO ix=-3.33 A

Reply
July 6, 2014

ezekeil aiyedun

This is a great discovery, I really appreciate your efforts on these solved problems with easy steps to understand.

Reply
July 15, 2014

mainak das

11/3=3.67

Reply
November 2, 2014

Doguscan Namal

After replacing the resistor, I found the solution of example as Ix1 = 1.388, Ix2 = 0, and therefore Ix has to be 1.388 right?

Reply
1. December 14, 2014
  
  Yaz
  
  Let’s see what others found!
  
  Reply
November 18, 2014

Luis Sanchez

I will like to view an example in superposition with three resistors in parallel with to voltage sources in the left resistor and other in the midle resistor to find the current in the right resistor first and in the midle resistor secondly.

Reply
February 24, 2015

John

0.14 A

Reply
March 11, 2015

Liwindi

I like the example given

Reply
March 27, 2015

Asos Engineer

Well done, That is great.

Reply
April 4, 2015

MB

I got my final current to be 12.5A?

I followed the same method as aboveand got 18A for the current from the first circuit and -6.5A for the second circuit, combining i got I=12.5A

Reply
1. April 4, 2015
  
  MB
  
  11.5lol
  
  Reply
May 8, 2015

richard A

Can I see an example of a linear circuit where the superposition theorem doesnt work?

Reply
1. May 8, 2015
  
  Yaz
  
  Superposition theorem works for all circuits. Of course, there should be more than one independent source to use the superposition.
  Sometimes we do not use the superposition simply because using other methods is easier.
  
  Reply
March 1, 2018

nick

if there was a resistor connected to the voltage source how would you do a source transformation on that to write a nodal matrix

Reply
June 29, 2018

John

Well done

Reply
July 17, 2018

Ron

Can i solve this circuit using the other theorem? (Source transformation)

Reply
September 2, 2018

V N Murthy

Ix1=1.4A and Ix2=0.01A and Ix=1.41A

Reply
September 22, 2018

Damon

2 01 A is answer

Reply
October 20, 2018

Mayuresh Joshi

Nice explanation 🙂

Reply
November 17, 2018

Zohaib Hassan

Keep it up..

Reply
June 12, 2019

Rakeshjinka

Ix1= 1.3888A
Ix2= 0.0000A
Hence Ix = Ix1 + Ix2 = 1.3888A

Reply
July 29, 2019

lokesh

1.38A

Reply
December 5, 2020

shweta

1.38 A for Ix 1 and 0A for so current is 1.38 A

Reply
December 17, 2020

YVLTanuja

Ans:Ix=2.339amps
When we kept 6ohms instead of 1ohm

Reply