# **URGENT** DC Superposition Question

#### sh312

Joined Dec 6, 2014
9
Would really appreciate if someone could provide a model answer for the attached question? I know that you have to short circuit both cells and work out the current flowing in each direction but then I just get all confusion.

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#### shteii01

Joined Feb 19, 2010
4,644
I got 0.428 Amperes.

#### sh312

Joined Dec 6, 2014
9
I got 0.428 Amperes.
Thank you, I have no idea how to do this, I'm attempting to crash an electrics module at university and have no idea how to do it :/ haha

#### shteii01

Joined Feb 19, 2010
4,644
You can use Mesh Current Method. You can use Node Voltage Method.

I did Superposition, then Mesh Current.

Step 1.
I shorted the 5 volt independent voltage source. That simplified the circuit into 6 volt independent voltage source and 3 resistors. Then I used Mesh Current Method to find current through RL.

Step 2.
I put 5 volt independent voltage source back into the circuit. I shorted 6 volt independent voltage source. That simplified the circuit into 5 volt independent voltage source and 3 resistors. Again, I used Mesh Current Method to find current through RL.

Step 3.
Take RL current from Step 1 and RL current from Step 2, add 'em.

#### sh312

Joined Dec 6, 2014
9
You can use Mesh Current Method. You can use Node Voltage Method.

I did Superposition, then Mesh Current.

Step 1.
I shorted the 5 volt independent voltage source. That simplified the circuit into 6 volt independent voltage source and 3 resistors. Then I used Mesh Current Method to find current through RL.

Step 2.
I put 5 volt independent voltage source back into the circuit. I shorted 6 volt independent voltage source. That simplified the circuit into 5 volt independent voltage source and 3 resistors. Again, I used Mesh Current Method to find current through RL.

Step 3.
Take RL current from Step 1 and RL current from Step 2, add 'em.
Thank you soo much for your help, I appreciate it however I still think I'm doing is wrong, this is how i've attempted it, please don't laugh haha!

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#### shteii01

Joined Feb 19, 2010
4,644
Thank you soo much for your help, I appreciate it however I still think I'm doing is wrong, this is how i've attempted it, please don't laugh haha!
RL is 10 Ohm resistor. They ask you to find current in RL. According to your drawing the current in RL is I3 and I5. Why the eF are you calculating I2 and I6?

#### sh312

Joined Dec 6, 2014
9
RL is 10 Ohm resistor. They ask you to find current in RL. According to your drawing the current in RL is I3 and I5. Why the eF are you calculating I2 and I6?
I have no idea, I've just accepted the fact I can't do it haha! I don't understand how to calculate RL once one of the voltage supplies is short circuited

#### shteii01

Joined Feb 19, 2010
4,644
I have no idea, I've just accepted the fact I can't do it haha! I don't understand how to calculate RL once one of the voltage supplies is short circuited
Replace one voltage source with short. Find voltage across RL. Apply Ohm's Law to RL to find current through RL.
Now do it again with the other voltage source.
Add the two currents.

#### sh312

Joined Dec 6, 2014
9
Replace one voltage source with short. Find voltage across RL. Apply Ohm's Law to RL to find current through RL.
Now do it again with the other voltage source.
Add the two currents.
is this correct? Thanks again!
My answer is 0.423A....

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#### shteii01

Joined Feb 19, 2010
4,644
is this correct? Thanks again!
My answer is 0.423A....
Yes. That is correct.