Resistor-Capacitor circuit

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.

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.

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 ]

[Old Man]

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

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

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.

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)

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

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

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 ..

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

Androcles.

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.

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