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7 votes

Why does Andy aka's calculation work in this case?

A mistake in your question states that 25 mA is flowing; it's 2.5 mA not 25 mA. Consider this first image (now your 2nd image due to you editing your post): - ...
Andy aka's user avatar
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Use of superposition principle for the inverting amplifier

I normally would just rush forward with the obvious voltage divider equation. But I'm going to assume, because you reference it, that you want each voltage source activated one at a time while all ...
jonk's user avatar
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6 votes

Is there a paradox in the superposition principle?

Sorry, I don't see any paradox. I don't think you understand how to use superposition. If there are \$N\$ independent sources you construct and analyze \$N\$ circuits, where each circuit has just one ...
Elliot Alderson's user avatar
6 votes

Intuition behind the superposition theorem

You already did describe it in the intuitive way. As to why it works. \$V=IR\$ describes a resistor and is linear. Linearity allows superposition by definition. The definition of superposition is ...
DKNguyen's user avatar
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Why are dependent sources not disabled with the superposition principle?

Two ways to think about this: If you disabled dependent sources when modeling the contribution of a dependent source to the circuit, they'd have no effect on the circuit. If you did another round of ...
The Photon's user avatar
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5 votes

Under which specific (formal) conditions can I apply the superposition theorem?

The superposition theorem works for any linear circuit. The problem with two voltage sources in parallel is that if their voltages are not identical then you have a nonsense circuit. Two different ...
Elliot Alderson's user avatar
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Is superposition theorem the only method to calculate the \$I_0\$ in this circuit?

i want to ask that why should use the superposition theorem to find the I0 in this circuit?must we use the superposition theorem to find the I0? No, unless you are instructed to do so. i mean ...
Andy aka's user avatar
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I get different results whenever I try to solve this circuit problem with superposition theorem. Can you please help?

Always re-draw your schematics. And keep in mind that you can select any one node and call it "the ground reference point." I decided to choose the following arrangement, but you are also ...
jonk's user avatar
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5 votes

What is wrong with this circuit analysis?

The circuit isn't as simple as it looks, as it contains an impossible situation--two voltage sources in parallel. The two voltage sources are both trying to assert a voltage between the same two ...
Hearth's user avatar
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Apply superposition principle to compute current in the 2 Ω resistor

Yes, that appears to be the right technique. You removed (open circuited the current sources) and correctly split the remaining driving current by the correct ratio. In other words, when removing the ...
Andy aka's user avatar
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How to Inject Noise into a DC Power Supply with a Function Generator

Usually you can make something like this work. Also read here: Bias-tee simulate this circuit – Schematic created using CircuitLab But part selection is a bit critical. In particular, the ...
tobalt's user avatar
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5 votes

How to solve for \$i\$ using superposition?

First off, redraw the schematic. In fact, you should get into the practice of redrawing every schematic. Even those that already look good. Just do it. And do it every time. The practice is worth its ...
periblepsis's user avatar
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4 votes

Solve for the current in the following circuit using superposition

The superposition theorem just says to replace remaining voltage sources with shorts and current sources with opens and evaluate. Then just sum everything up. I don't think any of your results are ...
jonk's user avatar
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4 votes

Use of superposition principle for the inverting amplifier

Applying the superposition principle, is the most simple method for finding the closed-loop gain of the inverter circuit. As long as the opamp is operated within its linear region we can set the ...
LvW's user avatar
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4 votes

Find the Norton equivalent of circuit using superposition

v is a fixed voltage source, nothing connected to it changes the voltage, so you can ignore the entire circuit left of v as well as R5.
Spehro Pefhany's user avatar
4 votes

Big confusion about the very fundamentals of circuit analysis

Another attempt is to calculate the Thevenin equivalents of the circuit.......we've proven that IR4 is positive; But this doesn't give us any direct information about IR3 and IR5. It pretty much does:...
Andy aka's user avatar
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Summing RMS values proof

The definition for root-mean-square (rms) voltage is: $$v_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}v^{2}dt}$$ If \$v=v_{1}+v_{2}\$ are all functions of time. Then: $$v_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}\...
RussellH's user avatar
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Why superposition theorem has not been executed here?

To apply superposition, you need to consider all the independent sources which are connected in the circuit, take their contributions and add them. Your circuit has only one independent source, the ...
sarthak's user avatar
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Why is the verification of Thevenin's theorem more accurate than that of Superposition theorem?

Why is Thevenin's theorem more accurate than the superposition theorem, experimentally? The last word in this question is the most important one. To verify the superposition theorem you measure your ...
le_top's user avatar
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Why does Andy aka's calculation work in this case?

Here is my method of analysis of this circuit with a floating V1 source at the input. From this diagram, we can see that \$V_P = I_{IN}R_3 = V_N\$ (due to negative feedback action). Additional we see ...
G36's user avatar
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3 votes

Why does Andy aka's calculation work in this case?

This answer is only for the problem with the superposition to calculate the currents. Superposition for two voltages is typically done with the cases \$(V1=0, V2)\$ and \$(V1, V2=0)\$. But that is not ...
JosefC's user avatar
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3 votes

What is wrong with this circuit analysis?

Update : Hypothesis : for "node analysis" ... "-- as long as no independent voltage sources form a loop .... ". It is the same limitation that occurs with a simulator! We can't ...
Antonio51's user avatar
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3 votes

What is wrong with this circuit analysis?

Other people gave explanations on why your model is flawed and hence produces weird results. I'll give you a mathematical explanation, instead. TL;DR: Simply put, you applied superimposition in a ...
LorenzoDonati4Ukraine-OnStrike's user avatar
3 votes
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(AC circuits) Why do I need to use the superposition theorem when dealing with two or more different sources with different frequencies?

You only need to use superposition if you are doing things in the frequency domain. Think about the equations you use for circuit elements in the frequency domain. Inductors are \$Z_L = j\omega L\$ ...
DKNguyen's user avatar
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Using the superposition theorem to solve for the output voltage

It seems that you should simplify the problem by combining U2 and U3 into what I've called U4: - Then, because of these words: - We've also assumed that R1 is much smaller than R2 We can make ...
Andy aka's user avatar
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3 votes

In superposition theorem, can I activate multiple sources together?

Think of superposition this way. You are not deactivating the other sources, you are setting them to zero output. They are still active, as in a low impedance voltage sources providing a known output ...
Neil_UK's user avatar
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Incorrect current assumed in Hayt Ex 5.3

The relationship between the dependent voltage source and the current must be maintained. If you swap the arrow direction for the current \$i_x\$ you also have to swap the sign on the dependent ...
Spehro Pefhany's user avatar
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How to solve for \$i\$ using superposition?

The simplest but most laborious method is the Superposition principle,Below I examine the various cases (see photo):
Franc's user avatar
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Voltage Mixer with 3 Resistors

Using superposition you can show that: (R1||R2||R3)/R1 = 1/2 and (R1||R2||R3)/R2 = 1/6 So if we set R1||R2||R3 = 10K (a more-or-less arbitrary choice) R1 = 20K R2 = 60K R3 = 1/(1/10K - 1/20K ...
Spehro Pefhany's user avatar

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