| Topic: |
Science > Physics |
| User: |
"novelo" |
| Date: |
22 Jul 2005 12:06:23 AM |
| Object: |
Circuits, capacitors, resistances, :S |
Hi there. I've wrote another message before about some problems that
are bothering me. I need to solve a guide to study for an exam, and
well, I'm very anxious because there are certain things I don't know
and eventhough I keep on trying to do them I cannot. So if anyone can
lighten me here I'll be truly thankful.
I've got a problem with resitance and capacitors.
R R
|-----/\/\/\----|---/\/\/\-----|
| | <---- |
| | I1 |
| -Q +Q | |
|------| |------| |
| C | -----
| | \ E ---
| I2 | / |
| | \ R |
| v / |
| | |
|---------------|--------------|
I think it is pretty understandable. Well, I've also been trying many
different things.
I've made some excercises explaing Kirchhoff's rule, but still I've
applyed it and got nothing but strange results.
If someone could show me THE PROCEDURE to do the following I'll be
very, but very thankful.
* What are the currents I1 and I2 in the figure?
* What is the electric charge Q charged in the capacitor.
I've tryed everything out, supossedly I've learned how to Circuits with
Resitances and Capacitors in the books, and I've handled all the
problems in my book, but this one seems impossible.
The answers to the excercise are:
I1= 2E/3R
12= E/3R
And
Q= CE/3
Thank you, thank you very much.
.
|
|
| User: "RobZompie" |
|
| Title: Re: Circuits, capacitors, resistances, :S |
22 Jul 2005 12:39:20 AM |
|
|
"novelo" <novelo@gmail.com> wrote in message
news:1122008783.208136.68660@g49g2000cwa.googlegroups.com...
Hi there. I've wrote another message before about some problems that
are bothering me. I need to solve a guide to study for an exam, and
well, I'm very anxious because there are certain things I don't know
and eventhough I keep on trying to do them I cannot. So if anyone can
lighten me here I'll be truly thankful.
I've got a problem with resitance and capacitors.
R R
|-----/\/\/\----|---/\/\/\-----|
| | <---- |
| | I1 |
| -Q +Q | |
|------| |------| |
| C | -----
| | \ E ---
| I2 | / |
| | \ R |
| v / |
| | |
|---------------|--------------|
I think it is pretty understandable. Well, I've also been trying many
different things.
I've made some excercises explaing Kirchhoff's rule, but still I've
applyed it and got nothing but strange results.
If someone could show me THE PROCEDURE to do the following I'll be
very, but very thankful.
* What are the currents I1 and I2 in the figure?
I1 comes out of the source, goes through R, then gets split-up and goes
equally through both the other Rs
That means that I2=1/2*I1.
So you have a simple voltage devsion of E by R and RIIR (parallel) => E2 is
1/3E
you should know how to convert parallel resistors into one value, then do
the voltage devider thing.
so Q=C*E/3 (since you know the voltage at C now)
and the rest is easy.....
* What is the electric charge Q charged in the capacitor.
I've tryed everything out, supossedly I've learned how to Circuits with
Resitances and Capacitors in the books, and I've handled all the
problems in my book, but this one seems impossible.
The answers to the excercise are:
I1= 2E/3R
12= E/3R
And
Q= CE/3
Thank you, thank you very much.
.
|
|
|
| User: "Helmut Wabnig " |
|
| Title: Re: Circuits, capacitors, resistances, :S |
22 Jul 2005 01:42:37 AM |
|
|
On Fri, 22 Jul 2005 00:39:20 -0500, "RobZompie" <invalid@spamless.com>
wrote:
"novelo" <novelo@gmail.com> wrote in message
news:1122008783.208136.68660@g49g2000cwa.googlegroups.com...
Hi there. I've wrote another message before about some problems that
are bothering me. I need to solve a guide to study for an exam, and
well, I'm very anxious because there are certain things I don't know
and eventhough I keep on trying to do them I cannot. So if anyone can
lighten me here I'll be truly thankful.
I've got a problem with resitance and capacitors.
R2 R1
|-----/\/\/\----|---/\/\/\-----|
| | <---- |
| | I1 |
| -Q +Q | |
|------| |------| |
| C | -----
| | \ E ---
| I2 | / |
| | \ R3 |
| v / |
| | |
|---------------|--------------|
I think it is pretty understandable. Well, I've also been trying many
different things.
I've made some excercises explaing Kirchhoff's rule, but still I've
applyed it and got nothing but strange results.
If someone could show me THE PROCEDURE to do the following I'll be
very, but very thankful.
* What are the currents I1 and I2 in the figure?
I1 comes out of the source, goes through R, then gets split-up and goes
equally through both the other Rs
That means that I2=1/2*I1.
So you have a simple voltage devsion of E by R and RIIR (parallel) => E2 is
1/3E
you should know how to convert parallel resistors into one value, then do
the voltage devider thing.
so Q=C*E/3 (since you know the voltage at C now)
and the rest is easy.....
Yes, because r2 and r3 are parallel and so is the cap
(added some numbers in the diagram, look)
w.
.
|
|
|
|
|
| User: "CWatters" |
|
| Title: Re: Circuits, capacitors, resistances, :S |
22 Jul 2005 01:33:17 AM |
|
|
"novelo" <novelo@gmail.com> wrote in message
news:1122008783.208136.68660@g49g2000cwa.googlegroups.com...
Assuming E is a DC voltage source and the resistors have equal value....
* What are the currents I1 and I2 in the figure?
Work on I1 first...
No DC current flows through the capacitor so for DC analysis you can remove
it. The two resistors in the lefthand half of the drawing are in parallel
and can be replaced by a single resistor of R//R = 0.5R
So now you have this circuit:
R
|---/\/\/\-----|
| <---- |
| I1 |
|
| |
| -----
\ E ---
/ |
\ 0.5 R |
/ |
| |
|--------------|
and so I1 = E/(R/2 + R) = E/1.5R = 2E/3R
Now calculate the voltage at the junction of the two resistors (Potential
divider). Note that this is also the voltage accross the capacitor in the
original circuit...
Vcap = E * 0.5R / (R+0.5R)
= E/3
Now going back to the original circuit you can work out the current I2
I2 = Vcap/R
= E/3R
* What is the electric charge Q charged in the capacitor.
The charge on a capacito is given by Q=VC or "Queen = Victoria Cross"
Q = E/3 * C or more neatly
= CE/3
.
|
|
|
|
| User: "Greg Neill" |
|
| Title: Re: Circuits, capacitors, resistances, :S |
22 Jul 2005 03:17:45 PM |
|
|
"novelo" <novelo@gmail.com> wrote in message
news:1122008783.208136.68660@g49g2000cwa.googlegroups.com...
Hi there. I've wrote another message before about some problems that
are bothering me. I need to solve a guide to study for an exam, and
well, I'm very anxious because there are certain things I don't know
and eventhough I keep on trying to do them I cannot. So if anyone can
lighten me here I'll be truly thankful.
I've got a problem with resitance and capacitors.
R R
|-----/\/\/\----|---/\/\/\-----|
| | <---- |
| | I1 |
| -Q +Q | |
|------| |------| |
| C | -----
| | \ E ---
| I2 | / |
| | \ R |
| v / |
| | |
|---------------|--------------|
I think it is pretty understandable. Well, I've also been trying many
different things.
I've made some excercises explaing Kirchhoff's rule, but still I've
applyed it and got nothing but strange results.
If someone could show me THE PROCEDURE to do the following I'll be
very, but very thankful.
* What are the currents I1 and I2 in the figure?
* What is the electric charge Q charged in the capacitor.
I've tryed everything out, supossedly I've learned how to Circuits with
Resitances and Capacitors in the books, and I've handled all the
problems in my book, but this one seems impossible.
The answers to the excercise are:
I1= 2E/3R
12= E/3R
And
Q= CE/3
Thank you, thank you very much.
At steady state for a DC-powered circuit, the capacitor will
"look" like an open circuit. You thus have an equivalent
circuit to solve that's fairly easy.
The net resistance seen by the voltage source will be:
Rnet = R + R||R = (3/2)R
So the current I1 is E/[(3/2R)] = 2E/3R
The voltage drop across the resistor with I1 is then I1*R = (2/3)E
This leaves E/3 across the parallel pair of resistors.
I think you can take it from here.
.
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