Find the Average Power absorbed by the 20k Resistor

[icesalmon]

Homework Statement

Find the Average Power absorbed by the 20k Resistor

Relevant Equations

V = I*R
A = Vout/Vin

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 V_A / ( 20k - j12k ) is omitted in the solution.

I_AC = I_AB + I_BC
V^- = V⁺ = V_B = V_S
( V_A-V_C )/(20k -j12k) = ( V_A - V_s )/(10k + j6k) + (V_s-V_c)/(2k+j4k)
V_c = V_ground = 0V
(V_A)/(20k-j12k) = (V_A-V_s)/(10k + j6k) + (Vs-Vc)/(2k+j4k)

 

[The Electrician]

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?

 

[DaveE]

I_AC = I_AB + I_BC

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.

 

[DaveE]

If I was solving this problem, I would just name all of those branch impedances, like Z_BC = 2KΩ + j4KΩ, etc. Then I would solve the problem in terms those named impedances. So, for example, the current through Z_BC would be Vs/Z_BC, 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.

 

[icesalmon]

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.
I_AB - I_AC = I_BC

 

[icesalmon]

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 Z_out as 20k - j12k?

 

[DaveE]

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.

 

[DaveE]

My equations are incorrect, so am I wrong in interpreting Z_out as 20k - j12k?

You have both defined and questioned the definition of Z_out in a single sentence, that doesn't make sense. Sure you can name 20k - j12k as Z_out.

 

[icesalmon]

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:
I_BC = I_AB
V_BC/(j4k + 2k) = V_AB/(10k + j6k)
Where Z_BC = j4k + 2k and Z_AB = 10k+ j6k

 

[DaveE]

ah okay, so:
I_BC = I_AB
V_BC/(j4k + 2k) = V_AB/(10k + j6k)
Where Z_BC = j4k + 2k and Z_AB = 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.

 

[icesalmon]

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 V_AB/Z_AB = V_AC/Z_AC?

 

[DaveE]

i'm still confused about what happens to the current when it reaches node A
Is V_AB/Z_AB = V_AC/Z_AC?

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.

 

[icesalmon]

nevermind, I think I got it. The purpose of an amplifier is to establish a larger voltage at the output (V_A in this case so V_A > V_B 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