Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

How do I find R1 current in this case with superposition theorem?

Status
Not open for further replies.

Sughyaliope

Newbie
Joined
Aug 7, 2022
Messages
1
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
7
1659845887832.png
 

Hi,

My solution:
(First, just to clarify: "Is", is not just a current, there also needs to be a voltage across "Is")
Vs is 13V, now R2 and R4 create a 6.5V source with a source resistance of 12.5 Ohms.
Combine this with R1 and you get a 6.5V source with 62.5 Ohms.

So we als know: I_R1 + I_R3 = 2A
Let's use Vx as the voltage at the bottom node (where R1, R3 and Is meet) wrt minus if the battery.
Then:
I_R3 = (13V - Vx) / 15 Ohms .... and
I_R1 = (6.5V - Vx) / 62.5 Ohms
This solves to Vx = - 1930/155 = -12.452V
And I_R1 = 0.3032A

Not 100% sure if my calculation is right.
What about a try with a simulation tool?

Klaus
 
Superposition theorem requires to calculate the effects of voltage and current source separately and add them. 0.3032 A is correct.
 
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top