# Mesh Analysis w/ Supermesh and "different" current direction

1. Oct 25, 2016

### eehelp150

1. The problem statement, all variables and given/known data
Find power absorbed by resistor, capacitor and inductor. Find the total complex power absorbed by all the loads.

2. Relevant equations
KVL/KCL

3. The attempt at a solution
I want to solve this using Mesh analysis. There is a supermesh where the 5A source is, so I start it like this:

How do I assign the mesh currents? I1 and I2 are going in different directions.

2. Oct 25, 2016

### cnh1995

You have been given reactances of the reactive elements while the sources seem to be dc. If you want to calculate the complex power, the sources must be ac. Is that the complete circuit diagram you have? Aren't the phase angles of the sources mentioned?

3. Oct 25, 2016

### eehelp150

Sorry about that, the phase angles are 0 degrees.

4. Oct 25, 2016

### Staff: Mentor

You can assign mesh currents in any direction you like so long as you write the mesh equations according to the resulting definitions. That is, when you write the equations pay attention to the defined current directions. The alternative is to just apply your wrote procedure and then interpret the results in terms of the currents you "really" want. That is, extract your $I_1$ and $I_2$ from the mesh currents after solving for the mesh currents; signs imply directions.

5. Oct 25, 2016

### eehelp150

-I1(10+j20) + 6V + I2(-j2) = 0
I1 + I2 = 5
Are these correct?

6. Oct 25, 2016

### Staff: Mentor

Yes, those look good!

7. Oct 25, 2016

### eehelp150

I1 = (-15-26i)/53
I2 = (280+26i)/53
Correct?

8. Oct 25, 2016

### Staff: Mentor

Yup.

9. Oct 25, 2016

### eehelp150

Finding the power absorbed by the inductor/resistor/capacitor is simply 6V * I? Where I = I2 for the capacitor and I1 for the resistor and inductor?

10. Oct 25, 2016

### Staff: Mentor

The 6 V voltage source is not across any of the individual components. Use the currents through the components and the component impedances. Remember the DC version P = I2R? There's an AC version of that. Hint: One of the I's in I2 should be the complex conjugate.

11. Oct 25, 2016

### eehelp150

So basically (I)(I*)(Z)

12. Oct 25, 2016

### Staff: Mentor

Correct. This works because with a little rearrangement we can see:

P = (I)(Z)(I*) = (V)(I*)

13. Oct 26, 2016

### eehelp150

Capacitor: -j2 * I2 * I2*
Resistor: 10 * I1 * I1*
Inductor: j20 * I1 * I1*

Correct?
Total would be everything added up?

14. Oct 26, 2016

### Staff: Mentor

Yes, that looks fine.

15. Oct 27, 2016

### eehelp150

I ended up getting these values, would you mind double checking?
Resistor = -1.61 + 2.8i
Capacitor = 10.37 - 55.3i
Inductor = -5.6 - 3.21i
What is the proper way to display these values? Converting to complex power(S = sqrt(P^2+Q^2))?

Just out of curiosity, would the voltage across resistor be -I1 * 10ohm because of the direction?

16. Oct 27, 2016

### Staff: Mentor

Can you show your calculations in detail? I'm immediately suspicious because the power you found for the resistor turned out to be complex; Resistors have no reactance and can only dissipate real power no matter if the current is complex or not. Inductors and capacitors have no resistance and cannot dissipate real power, yet your results show them both dissipating some real power.

For display of complex power in general I'm partial to the rectangular form so that I can immediately see the real and reactive parts. But some prefer to see the magnitude of the apparent power and the power factor. When in doubt provide both.
You need to specify how you will interpret a voltage before you even calculate it. That is, establish a reference point or orientation on your circuit diagram for interpretation of the voltage. Otherwise you can only talk about the magnitude of the potential change. When we write KVL equations, for example, we agree on interpreting the potential change in light of the arbitrarily decided current direction and direction of our "KVL walk" around the loop.
In the figure below, the voltage $V_R$ across the resistor would be positive in one interpretation and negative in the other. Yet it's the same potential drop. It's all in how you define the "measurement" on the circuit diagram.

17. Oct 27, 2016

### eehelp150

Forgot to use conjugate.
New values:
Resistor: 3.2W
Capacitor: -56.3 VAR
Inductor: 6.4 VAR

18. Oct 27, 2016

### Staff: Mentor

Yes, that's much better!

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