Parallel RC circuit equation

blacklionboy

Hi all,

Im currently trying to solve this problem for my college course and have become a bit stuck.

The question asks to find an equation for capacitor voltage (time dependent)

Im not sure where to start or to apply I= C(dv/dt).

If anyone could even give me the differential equation or just point me in the right direction it would be great.

Thanks

2011

blacklionboy wrote: »

Hi all
If anyone could even give me the differential equation or just point me in the right direction it would be great.

Thanks

DC I assume?
I haven't done this in a long time, but this may help to put you on the right track:

1) The initial voltage appearing across the capacitor will be equal to the voltage from the voltage source less the volt drop across the series resistor (R1). At the instant of switching on the parallel resistor (R2) will effectively be out of circuit as the capacitor will act as a dead short. At this time the current flowing in the circuit and through the capacitor = V / R1

2) The capacitor will charge and as it does it will cease to conduct. If you were to plot the current flowing through the capacitor against time it would look like this:



3) The final voltage across the capacitor will be equal to the voltage appearing across the parallel resistor.
The current flowing through the capacitor at this stage will be zero.
At this time the total current flowing in the circuit = V / (R1+R2)

Bruthal

The main value id use in any circuits these days is the charge time which is C(farads) x R(ohms) to give the time taken for the capacitor to charge to 2/3 full.

At instant of switch on, the voltage across the capacitor is 0v, and
the full supply voltage is across R1.

This page might be of some help, it gives the calculations for a capacitor in series with a resistor, which would be the same as your setup OP. The parallel resistor will just add more current to r1. So r1 will carry the same current as C +r2. C charging time will be r1 x C in seconds. So the calculations on the page can be done using the values of the series circuit of r1 and C.

Another similar one here, the capacitor voltage calculation for a given time is in the first paragraph in the blue box.

Like 2011, its been a fair while since i was doing many of the calculations on these things.

Irish_Elect_Eng

:-(

2011

Irish_Elect_Eng wrote: »

Vc = V


Seems to work.

I can't see how the applied voltage, V can be equal to the voltage across the capacitor, Vc @ T= infinity.
Remember there will be a volt drop across R1

Therefore @ T = infinity

Vc = V - (I1 x R1)

or Vc = I2 x R2

Bruthal

2011 wrote: »

I can't see how the applied voltage, V can be equal to the voltage across the capacitor, Vc @ T= infinity.
Remember there will be a volt drop across R1

Therefore @ T = infinity

Vc = V - (I1 x R1)

or Vc = I2 x R2

At time = 0s, the full supply voltage V will be across R1.

At T = infinity, the capacitor will be fully charged, and so there will be no current flow through the capacitor. Vc will = V

2011

robbie7730 wrote: »

At T = infinity, the capacitor will be fully charged, and so there will be no current flow through the capacitor.

Correct, as my graph above shows.
As the current flowing through the capacitor decreases the voltage across it increases.


However because the capacitor is effectively open circuit the circuit now consists of 2 resistors in series, R1 and R2.

∴
V = volt drop across R1 + volt drop across R2

But the capacitor is connected in parallel with R2

⇒
V = volt drop across R1 + Vc

Vc = V - (volt drop across R1)
= V - (I1 x R1)

pljudge321

The numerical answer is:

[LATEX]V_{c}(t)=V\frac{R_{2}}{R_{1}{plus}R_{2}}(1-e^{-\frac{t(R_{1}+R_{2})}{R_{1}R_{2}C}}) [/LATEX]

I got it by transforming into the laplace domain, solving the now trivial circuit and then transforming back into the time domain.

If college has thought me anything is that getting out of the time domain is usually the smartest thing you can do.

Edit: LaTex is being annoying, there should be a + between the R1 & R2 in the denominator of the fraction

Michael Collins

There's a few assumptions we must make before solving this circuit.

1. That the capacitor does not have a voltage on it at time t = 0
2. That the voltage source was at zero, and is suddenly turned on, with voltage V.

These are just two possible assumptions; the capacitor could be at any voltage initially, not just zero. Likewise the voltage source V, could be any waveform, not just a step function. Our assumptions above are probably what the examiner is asking. But the question, really, is poorly stated.

Anyway assuming the above, the solution is as pljudge321 has said. However, you don't have to use the Laplace Transform if you don't want to. Here's another method using basic circuit theory:



The circuit above has input voltage Vi and output voltage Vo(t). If the capacitor has zero voltage initially, and the input voltage is stepped from zero up to Vi at time t = 0 (as per our assumptions), the voltage across the capacitor as a function of time is given by:

[latex] \displaystyle v_{o}(t) = V_{i}\left(1-e^{{^{-t}/_{RC}}\right) [/latex]

This is a standard result for the above RC circuit (if anybody really wants, I can show how this can be dervied from first principles...).

The result above is one to remember, this is because many circuits - including the one in question - can be reduced to this form.

Our circuit for example, if we remove the capacitor, may be replaced by its Thévenin Equivalent:



This says that any linear circuit (i.e. black box) can be replaced by just an equivalent voltage source Vth, and equivalent resistance Rth as in the diagram above.

If we do this to the circuit blacklionboy has given without the capacitor, then add on the capacitor back on, it will be idential in form to the solution for the simple RC circuit above - just have different voltage and resistance values.

The Thévenin equivalent voltage is just the voltage of the circuit where the capacitor would be, assuming no current flowing out. By voltage division this is:

[latex] \displaystyle V_{th} = V \frac{R_2}{R_1 + R_2} [/latex].

Likewise, the Thévenin equivalent resistance is just the resistance looking into the circuit from where the capacitor would be, where the voltage source is set to zero (replaced by a short circuit). This is simply:

[latex] \displaystyle R_{th} = \frac{R_{1}R_{2}}{R_1 + R_2} = R_1 || R_2 [/latex]

that is, R1 in parallel with R2.

So now the answer is just

[latex] \displaystyle v_{o}(t) = V_{i}\left(1-e^{{^{-t}/_{RC}}\right) [/latex]

replaced with the values for this circuit, giving

[latex] \displaystyle v_{o}(t) = V_{th}\left(1-e^{{^{-t}/_{R_{th}C}}\right) [/latex]

[latex] \displaystyle v_{o}(t) = V \frac{R_2}{R_1 + R_2} \left(1-e^{{^{-t}/_{(R_{1} || R_{2}) C}}\right) [/latex].

TimeToShine

Michael how does that answer change if you're given an initial voltage across the capacitor? e.g Vc(0) = 5v or something. thanks

Bruthal

2011 wrote: »

⇒
V = volt drop across R1 + Vc

True, im going blind or something.
Vc = V/(R1+R2)xR2 when the capacitor is fully charged.

pljudge321

TimeToShine wrote: »

Michael how does that answer change if you're given an initial voltage across the capacitor? e.g Vc(0) = 5v or something. thanks

I haven't worked it out but my intuition would be that it would equal

[LATEX]\displaystyle
5(e^{-\frac{t(R_{1}+R_{2})}{CR_{1}R2}})+V\frac{R_{2}}{R_{1}+R_{2}}(1-e^{-\frac{t(R_{1}+R_{2})}{CR_{1}R2}})[/LATEX]

Michael Collins

TimeToShine wrote: »

Michael how does that answer change if you're given an initial voltage across the capacitor? e.g Vc(0) = 5v or something. thanks

pljudge321 is on the money again on this one. Since the circuit is linear you can treat the forced solution (due to the connected voltage source) and the transient solution (due to the initial voltage) separately, then add them to get the final answer due to both. We've already found the forced solution, so for the transient just set the voltage source to zero (short circuit), now we have a capacitor discharging through two parallel resistors, call this R. The equation of the circuit is given by Kirchoff's Current Law as

[latex] \displaystyle -\frac{v(t)}{R} = C \frac{\hbox{d}v}{\hbox{d}t} [/latex]

[latex] \displaystyle \frac{\hbox{d}v}{\hbox{d}t} = -\frac{1}{RC} v(t) [/latex]

This differential equation can be re-arranged and solved directly

[latex] \displaystyle \int_{v(0)}^{v(t)} \frac{\hbox{d}v}{v(t)} = \int_0^t -\frac{\hbox{d}t}{RC} [/latex]

[latex] \displaystyle \ln\left[\frac{v(t)}{v(0)}\right] = -\frac{t}{RC} [/latex]

[latex] \displaystyle v(t) = v(0)e^{^{-t}/_{RC}} [/latex]

Putting in the initial voltage and resistance values this finally becomes:

[latex] \displaystyle v(t) = 5e^{^{-t}/_{(R_1 || R_2)C}} [/latex]

Adding this to the forced solution at the end of my last post gets pljudge321's answer.