Engineering Please help me with this circuit question using the superposition theorem

AI Thread Summary
The discussion focuses on solving a circuit using the superposition theorem to find the current through three resistors. The first case involves activating a 5A current source while turning off a 25A source, resulting in a calculated current of 1.2A through one resistor. In the second case, the 25A source is activated, yielding a current of 15A through another resistor. The total current is then determined to be 16.2A, which aligns with the book's answers. The participants confirm the calculations and express gratitude for assistance.
snehil31
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Homework Statement
In the network shown in the figure, the two current sources provide I' and I" where I' + I" = I. Use superposition to obtain these currents.
Relevant Equations
Nodal analysis or Mesh analysis
I have tried many times to solve this network, but can't understand how to get current in each resistors by superposition theorem. Please help me to solve and find currents in each 3 resistors with solution.

Note:- The figure is attached below.
Screenshot 2021-02-20 171134.jpg
 
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Rules Violation: Providing a complete solution to another's homework is not permitted.
I haven't done circuit theory in years, but here goes nothing lol.

Just as a review, when you're using superposition theorem, you only "turn on" one source on at a time. In this case, you can split this circuit up into two cases:

5A Current Source = Let's call this Case #1.

25A Current Source = Let's call this Case #2Case #1: Only turn on "5A current source": Which means you turn off the 25A current source. So, the 25A current source will be an "open".

This means, you will be left with the 5A Current source in parallel with the 12 ohm resistor in parallel with the "38 ohm resistor"

So for this case, the 30 ohm resistor and the 8 ohm resistor have the same current because they are in series.

You can use the current divider equation in this case and find the current for the "38 ohm" ohm resistor, which then gives you the current for the 8 ohm resistor. This current can be called I'
I got 1.2 A for this current value.For Case #2: Only 25A current source on:

So for this case, the 5A current source is turned off. Which means the 5A current source becomes an "open".

This means that the 25A current source is now in parallel with the 30 ohm resistor and in parallel with the "20 ohm resistor" (i.e. because now the 8 ohm resistor is in series with the 12 ohm resistor).

The current through the 8 ohm resistor is the same as the current through the 12 ohm resistor, since they are in series.

You should be able to use current division to find the current through the "20 ohm resistor", which will give you the current through the 8 ohm resistor. This current value will be called I". I got 15A for this current value.

Your total "I" current value will be = I' + I"

My total I = I' + I" = 15A + 1.2A = 16.2A

Let me know what the final solution is to double check this work.
 
Last edited:
Thanks a lot @ammarb32 for the correct solution.
As per book the answers are this one: 1.2A, 15.0A & 16.2A. :smile:😇
 
snehil31 said:
Thanks a lot @ammarb32 for the correct solution.
As per book the answers are this one: 1.2A, 15.0A & 16.2A. :smile:😇

Awesome glad to hear. No worries best of luck!
 
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