Find the Average Power absorbed by the 20k Resistor

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    Average Power Resistor
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Discussion Overview

The discussion revolves around the analysis of an electrical circuit involving an operational amplifier (op-amp) and a 20kΩ resistor. Participants explore the implications of circuit components on gain, current flow, and voltage relationships, using node voltage analysis and circuit laws. The scope includes theoretical reasoning and technical explanations related to circuit behavior.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about the effect of the RC branch on the amplifier's gain, noting that certain equations appear omitted from the solution.
  • Others argue that if the op-amp is ideal, its output impedance is zero, suggesting that any impedance connected from the output to ground does not affect the circuit.
  • There are repeated references to Kirchhoff's laws, with participants emphasizing the need for voltages and currents to sum correctly in loops and nodes.
  • One participant proposes a method of naming branch impedances to simplify the problem-solving process, suggesting that this approach can help identify and correct mistakes more easily.
  • Some participants question the interpretation of the output impedance (Zout) and its implications for the circuit analysis.
  • There is a discussion about the current flow through resistors, with some asserting that the current through one resistor must also flow through another due to the infinite input impedance of the op-amp.
  • Participants explore the relationships between voltages and impedances at various nodes, with some expressing uncertainty about current behavior at node A.
  • One participant concludes that the purpose of the amplifier is to establish a larger output voltage, leading to a specific current flow through the circuit.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are multiple competing views regarding the impact of the op-amp's characteristics on the circuit and the interpretation of the equations involved. Some participants express confusion and seek clarification on various points, indicating that the discussion remains unresolved.

Contextual Notes

Limitations include potential misunderstandings of circuit laws, the interpretation of impedance definitions, and the assumptions made about the ideal behavior of the op-amp. There are also unresolved mathematical steps in the analysis.

icesalmon
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Homework Statement
Find the Average Power absorbed by the 20k Resistor
Relevant Equations
V = I*R
A = Vout/Vin
1607811359345.png

I wrote the following equations using node voltage, my issue is that I'm confused about why the RC branch at the output doesn't effect the value of the Gain for the amplifier; the equation involving VA / ( 20k - j12k ) is omitted in the solution.

IAC = IAB + IBC
V- = V+ = VB = VS
( VA-VC )/(20k -j12k) = ( VA - Vs )/(10k + j6k) + (Vs-Vc)/(2k+j4k)
Vc = Vground = 0V
(VA)/(20k-j12k) = (VA-Vs)/(10k + j6k) + (Vs-Vc)/(2k+j4k)
 
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If the opamp is ideal, which apparently it is, its output impedance is zero--it's an ideal voltage source so any impedance connected from the output (VA) to ground has no effect on the rest of the circuit.

Did you solve your equations and did they give the right answer?
 
Last edited:
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icesalmon said:
IAC = IAB + IBC
Nope. Try again.
KVL says that voltages added in a complete loop must sum to zero.
KCL says that currents flowing into a circuit node must sum to zero.
 
If I was solving this problem, I would just name all of those branch impedances, like ZBC = 2KΩ + j4KΩ, etc. Then I would solve the problem in terms those named impedances. So, for example, the current through ZBC would be Vs/ZBC, etc. I wouldn't actually calculate any complex arithmetic until I had the answer expressed in an equation with those named impedances. It doesn't have to be done this way, of course, but it's a style that I think is easier, particularly if you have to find and fix mistakes.
 
DaveE said:
Nope. Try again.
KVL says that voltages added in a complete loop must sum to zero.
KCL says that currents flowing into a circuit node must sum to zero.
Okay, I'm not sure which direction the currents actually move here.
IAB - IAC = IBC
 
The Electrician said:
If the opamp is ideal, which apparently it is, its output impedance is zero--it's an ideal voltage source so any impedance connected from the output (VA) to ground has no effect on the rest of the circuit.

Did you solve your equations and did they give the right answer?
My equations are incorrect, so am I wrong in interpreting Zout as 20k - j12k?
 
The op-amp has infinite input impedance, so no current flows into it's inputs. Therefore, the current through the 10K resistor HAS TO flow through the 2K resistor too.
 
icesalmon said:
My equations are incorrect, so am I wrong in interpreting Zout as 20k - j12k?
You have both defined and questioned the definition of Zout in a single sentence, that doesn't make sense. Sure you can name 20k - j12k as Zout.
 
DaveE said:
The op-amp has infinite input impedance, so no current flows into it's inputs. Therefore, the current through the 10K resistor HAS TO flow through the 2K resistor too.
ah okay, so:
IBC = IAB
VBC/(j4k + 2k) = VAB/(10k + j6k)
Where ZBC = j4k + 2k and ZAB = 10k+ j6k
 
  • #10
icesalmon said:
ah okay, so:
IBC = IAB
VBC/(j4k + 2k) = VAB/(10k + j6k)
Where ZBC = j4k + 2k and ZAB = 10k+ j6k
Yes, work from that. You will want to know what the op-amp output voltage is to determine what the 20K resistor power is.
 
  • #11
DaveE said:
Yes, work from that. You will want to know what the op-amp output voltage is to determine what the 20K resistor power is.
i'm still confused about what happens to the current when it reaches node A
Is VAB/ZAB = VAC/ZAC?
 
  • #12
icesalmon said:
i'm still confused about what happens to the current when it reaches node A
Is VAB/ZAB = VAC/ZAC?
The op-amp can source or sink current in order to make it's input voltages equal, because it has negative feedback. That is how op-amps work.
 
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  • #13
nevermind, I think I got it. The purpose of an amplifier is to establish a larger voltage at the output (VA in this case so VA > VB meaning current is driven from node A to node B and through to node C and also from node A to node C through the RC path. it doesn't travel through that feedback (AB) path from node B to A. sorry if this is confusing
 

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