Find total power in a circuit with two power supplies

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Discussion Overview

The discussion revolves around calculating the total power dissipated in a circuit with two power supplies using nodal analysis. Participants explore the application of circuit analysis techniques, including the formulation of equations for node voltages and the determination of currents from voltage sources.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory

Main Points Raised

  • Initial equations for node voltages VA and VB are proposed using nodal analysis.
  • Participants discuss the correct polarity of the 2 V source and its impact on the equations.
  • There is a suggestion to calculate currents produced by each voltage source to determine power dissipation.
  • One participant proposes using P=VI for calculating power from the voltage sources.
  • There is a discussion about the total resistance in the circuit and how to approach calculating it with multiple sources.
  • Participants express uncertainty about the correct current values and the implications of polarity on calculations.
  • One participant calculates the total power dissipated as 360 Watts based on their current values.

Areas of Agreement / Disagreement

Participants generally agree on the approach to use nodal analysis and the importance of polarity in calculations. However, there are competing views on the correct values of currents and how to calculate total power, indicating that the discussion remains unresolved in certain aspects.

Contextual Notes

Participants note the importance of checking polarities and the potential for errors in calculations due to assumptions made about current directions and voltage drops.

Who May Find This Useful

This discussion may be useful for students studying circuit analysis, particularly those working on homework related to nodal analysis and power calculations in electrical circuits.

Weaver
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Homework Statement



(a) Apply Nodal Analysis, writing down the equations which determine VA and VB, the voltages at nodes A and B [8]
(b) Find values for VA and VB. [4]
(c) By calculating the currents supplied by the 44 V and 2 V sources, or by any other means, calculate the total power dissipated by this circuit.
circuit.png

Homework Equations

The Attempt at a Solution


(a)[/B]
(Va-44)/4 + (Va- 0)/6 + (Va-VB)/1 = 0 ... Eq 1
(Vb -Va)/1 + (Vb -0)/3 + (VB-2)/2 = 0 ... Eq 2

(b)

Eq 1 ... x 12
3Va -132 + 2Va +12Va - 12 Vb = 0
17 Va - 12Vb = 132

Eq 2 ... x6
6Vb - 6Va + 2Vb + 3 Vb -6 = 0
11Vb - 6Va = 6

Eq 1 ... x 11
187Va - 132 Vb = 1452
Eq 1 ... x 12
-72 Va + 132 Vb = 66

115Va = 1518
Va = 13.2

=> Vb = 7.7 aprox

How do I go about doing (c)?
 
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Note the polarity of the 2 V source.
 
gneill said:
Note the polarity of the 2 V source.

Ok. So, should my (Vb - 2)/2 be (Vb +2)/2 ?
 
Conor_B said:
Ok. So, should my (Vb - 2)/2 be (Vb +2)/2 ?
Yes.
 
gneill said:
Yes.
Ok, thank you for pointing that out. So Va = 12.052 and Vb = 6.028 (approximately)

How do I go about doing C?
 
Conor_B said:
Ok, thank you for pointing that out. So Va = 12.052 and Vb = 6.028 (approximately)
Or, 12 V and 6 V exactly :smile:
How do I go about doing C?
Where is the power coming from? Where is it going?
 
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gneill said:
Where is the power coming from? Where is it going?

Haha, that's my problem. We haven't done to much on power in my lectures yet.

I'm assuming the power flows from the two voltage sources and is used up by the resistors as per joules law.

I'm just not sure how to get the current produced by each voltage source...

Do you calculate the voltage drop from 44V -> A and then divide by the resistance of 4 ?
And do something similar for the 2V source?
 
Conor_B said:
Haha, that's my problem. We haven't done to much on power in my lectures yet.

I'm assuming the power flows from the two voltage sources and is used up by the resistors as per joules law.
Correct.
I'm just not sure how to get the current produced by each voltage source...

Do you calculate the voltage drop from 44V -> A and then divide by the resistance of 4 ?
And do something similar for the 2V source?
Yup. In fact, that's very much like what you did to write the node equations, right? They're just KCL, summing the branch currents.
 
gneill said:
Correct.

Yup. In fact, that's very much like what you did to write the node equations, right? They're just KCL, summing the branch currents.

Ok. Cool. So the current in produced by 44V source is 8A and the 2V is 2A?

To get the total power dissipated by the circuit, do I get the total resistance in the circuit and multiply it by the sum of the two currents?
 
  • #10
Conor_B said:
Ok. Cool. So the current in produced by 44V source is 8A and the 2V is 2A?
One of those currents is correct :wink:
To get the total power dissipated by the circuit, do I get the total resistance in the circuit and multiply it by the sum of the two currents?
How would you find this "total resistance" when you have two sources? I suppose you could use the superposition theorem and do all the work twice...too much work for me! Or, you'd need the individual currents in every resistor. Again more work than is necessary.

Once you have the correct currents of the voltage sources, figure out how much power they each provide. That's where all the power is coming from, after all.
 
  • #11
gneill said:
One of those currents is correct :wink:

Ok, I think the 44V current is definitely correct. Is the 2V source -2A?

gneill said:
Once you have the correct currents of the voltage sources, figure out how much power they each provide. That's where all the power is coming from, after all.

Ah, so I just do P=VI for both of them and sum them together?
 
  • #12
Conor_B said:
Ok, I think the 44V current is definitely correct. Is the 2V source -2A?
Yes, the 44 V current is correct. What is your calculation for the 2 V source current? Did you pay attention to the polarity of the source?
Ah, so I just do P=VI for both of them and sum them together?
That's what I would do.
 
  • #13

Ah sugar. I forgot about that. So is it (6+2)/2 = 4A?
(We didn't talk about polarity too much in lectures)
 
  • #14
Conor_B said:
Ah sugar. I forgot about that. So is it (6+2)/2 = 4A?
(We didn't talk about polarity too much in lectures)
Yup. Always take a second to check polarities before you start writing equations. Make it a habit. They like to slip a reversed polarity in every once in a while to see if you're paying attention :wink: :smile:
 
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  • #15
gneill said:
Yup. Always take a second to check polarities before you start writing equations. Make it a habit. They like to slip a reversed polarity in every once in a while to see if you're paying attention :wink: :smile:

Thanks. I always make silly little mistakes like that. So, the total power dissipated by the circuit is 44(8) +2(4) = 360 Watts.

Thank you so much for your hep. My university doesn't post the solutions to past papers (which is where that question is from). It can be quiet annoying when you're just not quiet sure of an answer. So this has been a great help. I'll be sure to tell my class mates about this website :smile:
 
  • #16
You're welcome. Good luck in your studies!
 

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