Voltage Divider with multiple voltage sources

I am somewhat new to electronics (most of my knowledge is self-taught or using YouTube), but I came across a problem I have not been able to figure out. I need to find what Vout is at 1.0 second assuming ideal components and that all voltage sources were off initially.

So far, I have figured out that the voltage at the non-inverting input is v(1) = 6(1-e^(-1.0/(470010010^-6))) = 5.285 V. Additionally, I know voltage at the inverting output should be the same (5.285 V). But this is where I am lost because I am struggling to understand how all the components at the inverting output affect the voltage. Any help would be greatly appreciated.

Okay, so I tried using the superposition theorem as you described and created the following three equations: Va = Vout(3.3kΩ/43.3kΩ), Vb = V3(5kΩ/15kΩ) = -3(1/3) = -1, Vc = V4(10kΩ/15kΩ) = 8(2/3) = 5.333333, Va = 5.285V - (-1V + 5.333333V) = 0.952V, Vout = 0.952V(43.3kΩ/3.3kΩ) = 12.372V. This was not the right answer but I realized that the correct answer was actually my calculated value of Vout + Vb + Vc = 16.705V after using tinkercad simulations. Did I do something incorrect in my superposition calculations?

• Why not use a simulator - they are very popular these days. Oct 27, 2023 at 17:47
• Okay, I will try that. I am more interested the actual calculations for this circuit, though, and none of the simulators I have seen show the calculations. Oct 27, 2023 at 17:58
• @RonakPatel The calculations involved in a SPICE simulation wouldn't be very meaningful or educational to you, anyway, unless you were working on making a simulator yourself. Oct 27, 2023 at 18:03
• @RonakPatel can you edit the post and ask a specific question? Oct 27, 2023 at 18:19
• You know the voltage at the inverting input. You know the voltage at the other ends of R3 and R4. Therefore you know the current into the inverting node from R3 and R4. You know (or should know) that no current flows into the inverting node of the op-amp. Therefore you know the current through R2, and so the voltage across it, and so the voltage at Vout. That's the long-handed way to do it. Once you've done a few of these the long way, you'll appreciate gain equations involving the values of those resistors. However, I feel it's far better to do a few the long way to give you the incentive. Oct 27, 2023 at 18:23

Research the Superposition Theorem. This allows calculating Vout due to each input separately, then adding their contributions together.

Replace voltage sources with a short and current sources with an open. Leave the source that you’re working with intact.

For the input you’re working with replace V3 and V4 with a short. Then R3 is in parallel with R4.

Finish the calculation for Vout due to V2.

Then short V2 and restore V3. And so on.

Repeat for V4, then add them up.

You should be able to take it from here.

Update:

The TinkerCAD circuit has no relation to the original op-amp circuit shown. I suggest that you learn how op-amp circuits actually work.

Using superposition:

1. Replace V3 and V4 with a short circuit. This puts R3||R4 = 3.33kΩ.
2. $$\v_{IN+}=6e^{-t/\tau}\$$, so $$\v_{outV2}=\left(1+{40k \over 3.3k}\right) 6e^{-t/\tau}\$$ V
3. Replace V2 and V4 with a short circuit, so $$\v_{outV3}={-40k \over 10k}(-3)\$$ V
4. Replace V2 and V3 with a short circuit, so $$\v_{outV4}={-40k \over 5k}(8)\$$ V

Add these three valus together to get 16.7V

• Okay, so I tried using the superposition theorem as you described and created the following three equations: Va = Vout(3.3kΩ/43.3kΩ), Vb = V3(5kΩ/15kΩ) = -3(1/3) = -1, Vc = V4(10kΩ/15kΩ) = 8(2/3) = 5.333333, Va = 5.285V - (-1V + 5.333333V) = 0.952V, Vout = 0.952V(43.3kΩ/3.3kΩ) = 12.372V. This was not the right answer but I realized that the correct answer was actually my calculated value of Vout + Vb + Vc = 16.705V after using tinkercad simulations. Did I do something incorrect in my superposition calculations? Nov 1, 2023 at 22:00
• Hi @RonakPatel, No, the work isn't right. Paste your comment into the end of the question as an edit so that I can add the solution to my answer. Nov 2, 2023 at 0:19
• Okay, I added my comment to my original question as well as an image of the tinkercad simulation I made. Thanks for your help. Nov 4, 2023 at 15:48