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Question in DC circuit

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yousefsam

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hello all ;
i have a question , it seems easy but it has a trick the question is :
Find the current through each component using superposition .
the circuit in attachments , i solve the question and i found the value of current passing through each resistor is euqal 2A .
This answer is correct or wrong ,
plz help me ,
thanks .
 

Hi Yousefsam,

You should specify in which direction the current flows in each resistor
branch and you should also indicate the direction of the current flowing
through the 24 volt source. That was a smart answer, wasn't it.
You should make your homework yourself !

on1aag.
 

Open the current source and find the current. It is comming to be 2Amp in each branch in downwards direction.
Now, short the volatage source and connect the current source. Now find the currents in each resisttor.
The current conming in 7Ω resistance is 5Amp and current in 5ohm is 7amp.
So, total currents are
In 9Ω current is 2 amp downward
in 3Ω current is 2 Amp downward
in 5Ω current is 5 Amp downward
in 7Ω current is 7 amp downwards.
 

Ok guys :
this is the correct solution for the question :
i'm solved the question using mesh current , and i obtain the same result .
 

manojshukla said:
Open the current source and find the current. It is comming to be 2Amp in each branch in downwards direction.
Now, short the volatage source and connect the current source. Now find the currents in each resisttor.
The current conming in 7Ω resistance is 5Amp and current in 5ohm is 7amp.
So, total currents are
In 9Ω current is 2 amp downward
in 3Ω current is 2 Amp downward

in 5Ω current is 5 Amp downward
in 7Ω current is 7 amp downwards.
u didnt add the current due to the current source, it is 9 in 3 Ohm and 3 in 9 Ohm
 

Hi

Isn't:

7A downwards in the 9 ohm
3A upwards in the 5 ohm
5A upwards in the 3 ohm
9A downwards in the 7 ohm
 

nice and easy electronics superposition...

9ohm: 5A downwards
3ohm: 7A upwards
5ohm: 5A upwards
7ohm: 7A downwards

Im sure anyone with simulator(Proteus, electronics workbench. Matlab, PSpice, Tina/..etc can back this up.
 

If the circuit is linear (this is the case) then you can use superposition.

First you short the voltage generator and you find currents in each resistor due to the current generator.

After you open the current generator and you find the current in each resistor due to the voltage generator.

The current in each resistor when both the generators are active is the sum of the contribution calculated with only one generator active.
 

In 9Ω current is 2 amp downward
in 3Ω current is 2 Amp downward
in 5Ω current is 5 Amp downward
in 7Ω current is 7 amp downwards.
 

in 9Ω 5A
in 5Ω 5A
in 3Ω 7A
in 7Ω 7A

I`M sure

Added after 14 minutes:


in 9Ω 5A
in 5Ω 5A
in 3Ω 7A
in 7Ω 7A

I`M sure
64_1156759467.GIF
 

So we can disregard the voltage supply.....how?
 

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