# Setting up differential equations for voltage

• jerry fan
In summary, the conversation discusses solving for V1 and V2 as a function of time by using dV1/dt=I1/C1 and dV2/dt=I2/C2, and using circuit rules to express I1 and I2 as functions of V1 and V2. There is confusion about the equations and method of finding V1 and V2, but in the end it is determined that there needs to be a negative sign in the equation dV1/dt= -I1/C1 due to the relationship between current and voltage for capacitors.
jerry fan
Let V1 be the voltage across C1 and V2 be the voltage across C2. I want to solve for V1 and V2 as a function of time.

My idea was to use dV1/dt=I1/C1 and dV2/dt=I2/C2. Then using circuit rules i can express I1 and I2 as functions of V1 and V2 and substitute them into the previous diff eqs. Finally i throw the differential equations into mathematica and solve. However, I'm having issues with using circuit rules.

I1 is the current in C1 branch and points to the left
I2 is the current in C2 branch and points to the right
IRL is the current in IRL branch and points to the right

This is what I've come up with so far:
V1 - V2 + I2 * R2 =0
V1 = IRL * RL
I1 = IRL + I2

Are these equations and method of finding V1 and V2 as a function of time right? Thanks

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Hello Jerry, and welcome to PF !

Do you know how to solve this problem in case C1 is absent ?

BvU said:
Hello Jerry, and welcome to PF !

Do you know how to solve this problem in case C1 is absent ?

If C1 is absent, then V2=I*(R2+RL). Then dV2/dt=V2/(C2*(R2+RL))is that correct?

You wanted to solve for V1 and V2 and wrote down
V1 - V2 + I2 * C2 =0
V1 = IRL * RL
I1 = IRL + I2

so can you solve those ?

Now you add V2 = I * (R2+RL) but you don't tell me what I is. I suppose you mean I2 ?

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Could you also explain a little where V1 - V2 + I2 * C2 = 0 comes from ? Do you see that it can't be right ? The dimensions don't match.

BvU said:
You wanted to solve for V1 and V2 and wrote downso can you solve those ?

Now you add V2 = I * (R2+RL) but you don't tell me what I is. I suppose you mean I2 ?

---

Could you also explain a little where V1 - V2 + I2 * C2 = 0 comes from ? Do you see that it can't be right ? The dimensions don't match.
oh my bad i meant R2. I dindt label I because its could be labeled IRL or I2

So by now we are at the following picture, right ?

In that case I wholeheartedly agree with what you wrote in post #1: it's correct.

 correction ! dV1/dt=I1/C1 isn't correct !

You now have five equations with five unknowns. You might remove a few but that's up to you.
Sure you want to "throw it all into mathematica" ?

If C1 is absent, then V2=I*(R2+RL). Then dV2/dt=V2/(C2*(R2+RL))is that correct?
Yes. Would you need mathematica in that case , or could you solve it on your own ?

 same correction: needs a minus sign

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BvU said:
So by now we are at the following picture, right ?

View attachment 87552

In that case I wholeheartedly agree with what you wrote in post #1: it's correct.

You now have five equations with five unknowns. You might remove a few but that's up to you.
Sure you want to "throw it all into mathematica" ?

Yes. Would you need mathematica in that case , or could you solve it on your own ?
Yes, that exactly what I had in mind. Thanks for the confirmation. I havn't had to use circuits in a while so I am pretty rusty. I can solve by hand but i have to use them later so i decided that might as well get mathematica to solve for me.

Check the signs. I edited and re-edited #6.

There are only two degrees of freedom in the problem; I suggest you work it out so far that you only solve two equations with two unknowns ...

BvU said:
Check the signs. I edited and re-edited #6.

There are only two degrees of freedom in the problem; I suggest you work it out so far that you only solve two equations with two unknowns ...
in #6 what is wrong with dV1/dt=I1/C1? i solved the equations by hand and got
I1=((R2 - RL)*V1 + RL*V2) / (R2 RL)
and
I2= (-V1+V2)/R2

At this point I cannot substitute I1 and I2 into the equation dV/dt=I/C?

It should be dV1/dt= ## \bf - ## I1/C1

BvU said:
It should be dV1/dt= ## \bf - ## I1/C1
Can you explain why there needs to be a negative sign there?

For the capacitor you have Q = CV so I = dQ/dt = C dV/dt : A positive I means the Q increases hence the V increases.

Your definition of I1 let's Q decrease if I1 is positive

## 1. What is the meaning of "differential equations" in terms of voltage?

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. In the context of voltage, they are used to model the changes in voltage over time.

## 2. Why are differential equations necessary when working with voltage?

Differential equations are necessary when working with voltage because they allow us to model the behavior of electrical circuits and systems. By setting up differential equations, we can predict how voltage will change over time and use this information to design and analyze circuits.

## 3. How do you set up differential equations for voltage?

To set up differential equations for voltage, you must first identify the components in the circuit and their relationships. Then, you can use Kirchhoff's voltage law and Ohm's law to write equations for each component. Finally, you can combine these equations and apply any necessary initial or boundary conditions to create a system of differential equations.

## 4. Are there any techniques or strategies for solving differential equations for voltage?

There are several techniques for solving differential equations for voltage, including separation of variables, substitution, and using Laplace transforms. It is important to choose the appropriate technique based on the complexity of the equations and the desired level of accuracy.

## 5. How can differential equations for voltage be applied in real-world scenarios?

Differential equations for voltage have a wide range of applications in various fields such as electronics, power systems, and telecommunications. They are used to design and analyze electrical circuits, predict the behavior of complex systems, and optimize the performance of devices and networks. They are also essential in understanding and troubleshooting issues with voltage fluctuations and stability in real-world scenarios.

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