# Resistor-Capacitor circuit

Discussion in 'Electronic Basics' started by Joachim, Nov 15, 2005.

1. ### JoachimGuest

How can I solve the following problem: "Show that the voltage Vc (t)
across the capacitor as a function of time goes as 4.125(1 -
exp(-t/0.1375)) volts. When the steady state has been reached, what is
the power dissipated in each resistor and the energy stored in the
capacitor?"

Circuit diagram: http://krebs.uk.com/rc.jpg

I know how to do this for a normal RC circuit, without the 22kOhm
resistor in parallel around the capacitor. I think I need to somehow
use Kirchhoff laws to figure this out, but I could really use a hand
getting started.

Joachim

Joachim, Nov 15, 2005

2. ### AJWGuest

Try drawing the circuit a little differently.

Move the cap to the right.

Now, ask yourself questions about the source the cap is seeing: In
effect it's a voltage source (how big?) and a series resistance (how
big?).

The diagram you supplied is a neat way of confusing the issue.

AJW, Nov 15, 2005

3. ### AJWGuest

Try drawing the circuit a little differently.

Move the cap to the right.

Now, ask yourself questions about the source the cap is seeing: In
effect it's a voltage source (how big?) and a series resistance (how
big?).

The diagram you supplied is a neat way of confusing the issue.

AJW, Nov 15, 2005
4. ### Christian S.Guest

On 15 Nov 2005 11:13:58 -0800, "Joachim" <> wrote:

>How can I solve the following problem: "Show that the voltage Vc (t)
>across the capacitor as a function of time goes as 4.125(1 -
>exp(-t/0.1375)) volts. When the steady state has been reached, what is
>the power dissipated in each resistor and the energy stored in the
>capacitor?"
>
>Circuit diagram: http://krebs.uk.com/rc.jpg
>
>I know how to do this for a normal RC circuit, without the 22kOhm
>resistor in parallel around the capacitor. I think I need to somehow
>use Kirchhoff laws to figure this out, but I could really use a hand
>getting started.
>
>
>Joachim

There is only one node in this circuit that is not fixed.
Let's call this node A and the node voltage va.
Apply KCL at this node: -(6V - va)/10k + va/22k + 20uF* dva/dt = 0

Solve the differential equation and you are done.

Chris

Christian S., Nov 15, 2005
5. ### Old ManGuest

"Joachim" <> wrote in message
news:...
> How can I solve the following problem: "Show that the voltage Vc (t)
> across the capacitor as a function of time goes as 4.125(1 -
> exp(-t/0.1375)) volts. When the steady state has been reached, what is
> the power dissipated in each resistor and the energy stored in the
> capacitor?"
>
> Circuit diagram: http://krebs.uk.com/rc.jpg
>
> I know how to do this for a normal RC circuit, without the 22kOhm
> resistor in parallel around the capacitor. I think I need to somehow
> use Kirchhoff laws to figure this out, but I could really use a hand
> getting started.

The two resistors form a voltage divider such that the
"steady state" voltage, V_cmax, accross the capacitor
is given by

V_cmax = V_supply * [ R_parallel / ( R_series + R_parallel )]

whereof V_cmax = 4.125 V

The steady state energy, E, stored in the capacitor is

E = ( C / 2 ) * ( V_cmax )^2

The steady state power dissipation in the resistors are
those of the voltage divider. In the steady state, the
capacitor draws no current.

The RC time constant of the circuit is

R*C = 0.1375 seconds

Thus, with C = 20 uF, R = 6875 Ohms which is the
resistance of R_series and R_parallel when hooked in
parallel

R = R_series * R_parallel / [ R_series + R_parallel ]

> Joachim

[Old Man]

Old Man, Nov 15, 2005
6. ### CWattersGuest

"Joachim" <> wrote in message
news:...
> When the steady state has been reached, what is
> the power dissipated in each resistor and the energy stored in the
> capacitor?"

Hint: Steady state = DC and capacitors behave like an open circuits for
DC... so remove the capacitor for now...

Then calculate the steady state current flowing...

I = V/R
= 6/(10,000+22,000)
= 187u5A

Power in 22K..
= I^2 x R
= 773uW

Power in 10K
= I^2 x R
= 352uW

Voltage on the 22K Res...

= I x 22K

or because it's a classic potential divider...

= 6 x 22/(10+22)

= 4.125V

Now put the capacitor back...

Voltage on capacitor is also 4.125V

Energy stored in a capacitor

= 0.5 C V^2

= 0.5 x 20E-6 x 4.125^2

= 170uJ

Assuming I haven't made a typo.

In the real world you might want to check the leakage current through the
capacitor can be ignored.

CWatters, Nov 15, 2005
7. ### Black KnightGuest

"Joachim" <> wrote in message
news:...
> How can I solve the following problem: "Show that the voltage Vc (t)
> across the capacitor as a function of time goes as 4.125(1 -
> exp(-t/0.1375)) volts. When the steady state has been reached, what is
> the power dissipated in each resistor and the energy stored in the
> capacitor?"
>
> Circuit diagram: http://krebs.uk.com/rc.jpg
>
> I know how to do this for a normal RC circuit, without the 22kOhm
> resistor in parallel around the capacitor. I think I need to somehow
> use Kirchhoff laws to figure this out, but I could really use a hand
> getting started.
>
>
> Joachim

Whoa.... is this a homework problem? The only person who
should be doing your homework is you, but I don't mind helping.

Break the problem down.

The 10k and 22k resistors form a voltage divider.
What is the final voltage across the capacitor?
(Pretend the capacitor isn't there.)

When the current is first applied (switch not shown)
what was the voltage across the capacitor?

At switch on (t = 0), the capacitor is effectively a short circuit.
At t = infinity, it is effectively an open circuit.

6 * 10/22 = ?
Looks like your 4.125 isn't quite right. Maybe that's the problem.

What is exp(-t) when t is large?
You can use a calculator for that.
Hint: On Microsoft Windows calculator, use Inv (the check box) ln
(the natural logarithm function)

What does the 0.1375 RC time constant represent?

Hope that helps.
Androcles

Black Knight, Nov 15, 2005
8. ### JoachimGuest

Thanks for the help guys, I'm slowly getting to grips with this
textbook.

Joachim, Nov 15, 2005
9. ### Bob MonsenGuest

On Tue, 15 Nov 2005 11:13:58 -0800, Joachim wrote:

> How can I solve the following problem: "Show that the voltage Vc (t)
> across the capacitor as a function of time goes as 4.125(1 -
> exp(-t/0.1375)) volts. When the steady state has been reached, what is
> the power dissipated in each resistor and the energy stored in the
> capacitor?"
>
> Circuit diagram: http://krebs.uk.com/rc.jpg
>
> I know how to do this for a normal RC circuit, without the 22kOhm
> resistor in parallel around the capacitor. I think I need to somehow use
> Kirchhoff laws to figure this out, but I could really use a hand getting
> started.

One way to look at it is that the cap is being fed from a voltage source
that has a series resistance of 10k || 22k, and a voltage that is equal to
the voltage divider voltage.

Notice that 6*22/(22+32) = 4.125, and that (10k||22k)*20e-6 = 0.1375.

If your textbook has mentioned Thevenin's theorem, that is simply the more
formal way to do this. It states that if you have a two-terminal
passive element in a circuit, you can find another circuit that consists
of the device, a single resistor, and a single current source,
all in series. The value of the voltage source is the thevenin equivalent
voltage, and the value of the resistor is the thevenin equivalent
resistance. The method to do this is trivial:

1) Remove the component you are analyzing from the schematic (in this case
C), and determine the voltage between the place where the leads previously
connected. In this case, it is V = 6 * (22/(10+22)). That is the thevenin
equivalent voltage.

2) Now, in that modified circuit, replace the voltage source with a wire,
and compute the resistance between those same two points. In this case, it
is 10k || 22k = 6.875k. This is the thevenin equivalent resistance.

Your circuit is then equivalent to

4.125V ------[6.875k ohms]------[20uF]-----GND

from the point of view of the cap, which is far easier to analyze.

Regarding the resistors, when the cap is in its 'steady state', it
effectively disappears from the circuit. So, use the equation for power
given voltage and resistance: (6-4.125)^2/10000 and 4.126^2/22000.

---
Regards,
Bob Monsen

Nature does not at once disclose all Her mysteries. - Lucius Seneca (Roman
philosopher)

Bob Monsen, Nov 16, 2005
10. ### CWattersGuest

"Black Knight" <> wrote in message
news:qCtef.8033\$...
> Whoa.... is this a homework problem? The only person who
> should be doing your homework is you, but I don't mind helping.
>
> Break the problem down.
>
> The 10k and 22k resistors form a voltage divider.
> What is the final voltage across the capacitor?
> (Pretend the capacitor isn't there.)
>
> When the current is first applied (switch not shown)
> what was the voltage across the capacitor?
>
> At switch on (t = 0), the capacitor is effectively a short circuit.
> At t = infinity, it is effectively an open circuit.
>
> 6 * 10/22 = ?
> Looks like your 4.125 isn't quite right. Maybe that's the problem.

Nope you made an error. The equation is ..

6 * 10/(10+22) = 4.125

CWatters, Nov 16, 2005
11. ### CWattersGuest

"CWatters" <> wrote in message
news:w7Cef.49826\$-ops.be...
>
> Nope you made an error. The equation is ..
>
> 6 * 10/(10+22) = 4.125
>

Now you got me making mistakes!...

The voltage drop accross the 22K is...

6 * 22/(10+22) = 4.125

The voltage drop accross the 10K is

6 * 10/(10+22) = 1.875

CWatters, Nov 16, 2005
12. ### Black KnightGuest

"CWatters" <> wrote in message
news:w7Cef.49826\$-ops.be...
>
> "Black Knight" <> wrote in message
> news:qCtef.8033\$...
> > Whoa.... is this a homework problem? The only person who
>> should be doing your homework is you, but I don't mind helping.
>>
>> Break the problem down.
>>
>> The 10k and 22k resistors form a voltage divider.
>> What is the final voltage across the capacitor?
>> (Pretend the capacitor isn't there.)
>>
>> When the current is first applied (switch not shown)
>> what was the voltage across the capacitor?
>>
>> At switch on (t = 0), the capacitor is effectively a short circuit.
>> At t = infinity, it is effectively an open circuit.
>>
>> 6 * 10/22 = ?
>> Looks like your 4.125 isn't quite right. Maybe that's the problem.

>
> Nope you made an error.

Yep... My apologies. I realised it right after I posted, so I'll have
to eat it. I 'fess up, I made a boo-boo.My granddaughter called
as I was about to re-read what I'd written, so I posted in haste.

The equation is ..
>
> 6 * 10/(10+22) = 4.125

Of course. I didn't engage brain before opening my big mouth,
so call me an idiot (this time).

Androcles.

Black Knight, Nov 16, 2005
13. ### Black KnightGuest

"CWatters" <> wrote in message
news:laCef.49827\$-ops.be...
>
> "CWatters" <> wrote in message
> news:w7Cef.49826\$-ops.be...
> >
>> Nope you made an error. The equation is ..
>>
>> 6 * 10/(10+22) = 4.125
>>

>
> Now you got me making mistakes!...
>
> The voltage drop accross the 22K is...
>
> 6 * 22/(10+22) = 4.125
>
> The voltage drop accross the 10K is
>
> 6 * 10/(10+22) = 1.875

6 - 4.125 = 1.875

BTW, the speed of light from
(AB+BA)/( t'A-tA) = c =0 <> 300,000,000 meters/second.

In SR, c = 0.
That's the biggest boo-boo in the history of physics, has cost
billions of m.o.n.e.y, so I don't feel all that bad.
Androcles.

Black Knight, Nov 16, 2005
14. ### John C. PolasekGuest

On 15 Nov 2005 11:13:58 -0800, "Joachim" <> wrote:

>How can I solve the following problem: "Show that the voltage Vc (t)
>across the capacitor as a function of time goes as 4.125(1 -
>exp(-t/0.1375)) volts. When the steady state has been reached, what is
>the power dissipated in each resistor and the energy stored in the
>capacitor?"
>
>Circuit diagram: http://krebs.uk.com/rc.jpg
>
>I know how to do this for a normal RC circuit, without the 22kOhm
>resistor in parallel around the capacitor. I think I need to somehow
>use Kirchhoff laws to figure this out, but I could really use a hand
>getting started.
>
>
>Joachim

Look up Thevenin's theorem. It says that your circuit can be replaced
by one with a series resistor equal to the two of them in parallel and
a battery voltage that is reduced by the voltage divider effect. Old
man had the right idea but Thevenin's is simpler.

John Polasek

John C. Polasek, Nov 16, 2005