Find total power in a circuit with two power supplies

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SUMMARY

This discussion focuses on calculating the total power dissipated in a circuit with two power supplies using Nodal Analysis. The voltages at nodes A and B were determined to be approximately 12.052V and 6.028V, respectively. The total power dissipated was calculated using the formula P=VI, resulting in a total of 360 Watts, derived from the currents produced by the 44V and 2V sources. Key insights include the importance of checking the polarity of voltage sources and applying Kirchhoff's Current Law (KCL) for accurate current calculations.

PREREQUISITES
  • Nodal Analysis for circuit analysis
  • Kirchhoff's Current Law (KCL)
  • Understanding of voltage sources and their polarities
  • Basic power calculations using P=VI
NEXT STEPS
  • Learn advanced Nodal Analysis techniques for complex circuits
  • Study the Superposition Theorem for circuits with multiple sources
  • Explore the implications of polarity in circuit design and analysis
  • Review Joule's Law and its applications in electrical engineering
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Electrical engineering students, circuit designers, and anyone involved in analyzing power systems will benefit from this discussion.

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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