Would really appreciate if you guys can help me with my homework Thanks in advance!


MOD NOTE: Deleted all but first problem. Showed problem within post body.

 

Three things:

1) You should only post one problem in a thread. Otherwise the thread becomes total chaos as different people respond to different problems.

2) You need to show YOUR best effort to solve YOUR homework problems. We will not work them for you and we need to see where YOU are going wrong in order to help you move forward.

3) Don't go and start four different threads on four different problems. Pick one. Work it through. Apply what you have learned to the next one before starting a thread on it and showing your new best attempt.

I'm going to arbitrarily delete all but your first problem, but you still need to post your best attempt to get started on it.

 

So with superposition you solve for the currents one source at a time, then algebraically add all the currents at a point to get the resultant current flowing from all sources at that point.

 

Sry i didnt know the rules here but i answered my 2nd and 3rd problem tho! Ty

 

For the 123 solution:

With I1=4.1A the voltage drop across the 6Ω resistor is 4.1x6=24.6V. Then the voltage drop across the 3Ω resistor must be 30-24.6=5.4V and I2=5.4÷3=1.8A.

But, if the voltage drop across the 3Ω resistor is 5.4V then the voltage drop across the 9Ω resistor must be 15V-5.4V=9.6V so that would make I2=9.6V÷9Ω=1.07A, yet your calculation shows I2=1.83A!

Did I do that right? Or did you do yours wrong?

 

Can you give me your solution please im probably wrong because when i calculated the voltage drop for 3Ω using superposition the result should be 10.91V!

 

Can you give me your solution ... the voltage drop for 3Ω using superposition the result should be 10.91V

This must be the 123 problem. Nodal analysis gives the voltage drop for 3Ω resistor as 120/11 V.

 

So the answer is 120/11 V? For the voltage drop?

 

So the answer is 120/11 V? For the voltage drop?

It is the exact answer, not a floating point approximation. 120/11=10.90909091