# Need help with 2 capacitor 2 resistor circuit

• Engineering

## Homework Statement

Find an expression for Vo, the voltage across the resistor. Hint: find V1 and V2 first.

## Homework Equations

i_c = C(dV/dt)
V_c = 1/C integral( i dt )

## The Attempt at a Solution

I think at t = 0, the voltage source turns on so

@ t < 0 :

V1(0-) = 0 V2(0-) = 0

@ t > 0 :

V1(0+) = V1(0-) = 0 V2(0+) = V2(0-) = 0
V1(final) = 5 V2(final) = 5

Next, I guess I'm supposed to write nodal equations:

I have:

5/1 = 0.5 (dV1/dt) + 0.333 (dV2/dt)

(V1 - V2)/1 = 0.333 (dV2/dt)

V0 = V1 - V2

I'm not sure how to proceed from here. The book examples say I need to get a characteristic equation somehow, but those have inductors in them. Would appreciate any help. btw, the answer is supposed to be Vo = 2(e^-t - e^-6t)

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NascentOxygen
Staff Emeritus
Next, I guess I'm supposed to write nodal equations:

I have:

5/1 = 0.5 (dV1/dt) + 0.333 (dV2/dt)
Re-check this equation.

5/1 = 0.5 (dV1/dt) + (V1 - V2)/1

I thought the current through the second resistor should be the same as through the 1/3 F capacitor...?

NascentOxygen
Staff Emeritus

5/1 = 0.5 (dV1/dt) + (V1 - V2)/1
It's the 5/1 that I can't explain.

I just did a source transformation to find the current going into node V1.

NascentOxygen
Staff Emeritus
Try it without that.

Then it would be 0 = 0.5 (dV1/dt) + 0.333 (dV2/dt) ?
I'm not sure what you meant.

NascentOxygen
Staff Emeritus
Keep it as a voltage source, and write the current through that first 1 ohm resistor.

So:

(V1 - 5)/1 = 0.5 (dV1/dt) + 0.333 (dV2/dt)

I tried that at first but I wasn't sure what to do with all the variables.

NascentOxygen
Staff Emeritus
So:

(V1 - 5)/1 = 0.5 (dV1/dt) + 0.333 (dV2/dt)

I tried that at first but I wasn't sure what to do with all the variables.
I think it's (5 - V1)/1 = ....

That's an equation in two variables. You need to find another relation:-

Earlier, you also wrote:
(V1 - V2)/1 = ⅓ dV2/dt
⇔ V1 = ⅓ dV2/dt + V2

Differentiate both sides to find dV1/dt then substitute for V1 and dV1/dt in your equation, and you have an equation in only one variable. You have a second order DE to solve.

Okay, now I have:

V1 = ⅓ dV2/dt + V2
dV1/dt = ⅓ d²V2/dt² + dV2/dt

~~

⅓ dV2/dt + V2 - 5 + ½(⅓ dV2/dt + dV2/dt) + ⅓ dV2/dt = 0
⅓ dV2/dt + V2 - 5 + 1/6 d²V2/dt² + ½ dV2/dt + ⅓ dV2/dt = 0
1/6 d²V2/dt² + 7/6 dV2/dt + V2 - 5 = 0

I wasn't sure what to do with that -5 for the characteristic equation so I just dropped it.

so I ended up with:
1/6s² + 7/6s + 1 = 0
s² + s + 6 = 0
(s + 1)(s + 6)
s = -1, -6
~~

So does V0 = Ae-t + Be-6t ?

I'm also unsure if my initial and final conditions are correct.

FBS,

So does V0 = Ae-t + Be-6t ?
No, the equation is incomplete. You should not have dropped the -5. You are treating the equation like it is homogeneous. Read up on linear nonhomogeneous equations with constant coefficients.

So does V0 = Ae-t + Be-6t ?
No, you are missing a term because you did not take the nonhomogeneous term into account. By the way, the supposed answer in post #1 is wrong because it goes to zero when time goes to infinity.

Ratch

NascentOxygen
Staff Emeritus
s² + s + 6 = 0
(s + 1)(s + 6)
s = -1, -6
~~

So does V0 = Ae-t + Be-6t ?
You are not up to finding V0 yet. At the moment you are solving a D.E. to determine V2. Then you'll solve another D.E. to determine V1. Then subtract those to find V0. That must be what the examiner had in mind, anyway, in light of the hint to first find V1 and V2. (*)

But before going farther, suppose you were dealing with a single R-C stage, the 1Ω and 0.5F. What would be the expression for V1 in this simpler case? Can you write V1(t) for this simple case by inspection?

(*) I think I see a bit of a shortcut. In an earlier post, you wrote:
Okay, now I have:

V1 = ⅓ dV2/dt + V2
We ultimately are seeking V1-V2, and this equation says V1-V2 = ⅓ dV2/dt
so once you have determined V2, if you differentiate it and divide by 3, you'll have V0.

Last edited:
NascentOxygen,

You are not up to finding V0 yet. At the moment you are solving a D.E. to determine V2. Then you'll solve another D.E. to determine V1.
The OP only needs to solve the DE he already calculated, which is "6 d²V2/dt² + 7/6 dV2/dt + V2 - 5 = 0". This is readily solved by using standard methods for nonhomogeneous linear equatons with constant coefficients. He did not and does not have to calculate V1 to solve for Vo.

Ratch