# Circuit help - resistor and capacitors... time too!?!?

#### Supervisor

Joined Jan 17, 2009
32
Hello,

I've just learned about the basics of circuits but then my teacher gives us a take home problem that contains resistors and capacitors...
On top of that, the problem also involves using specific seconds to measure the values we need.

I don't know if I missed a day for the notes on this, but I have absolutely no idea how to do this!

The circuit can be found here:

Link (if the above doesn't show): http://www.fileden.com/files/2008/9/4/2081882//Circuit.bmp

Help with just this one question would be awesome and then maybe I can get the rest!

"... At t = 0s switch one (1) is closed but switch three (3) remains open
a) What current flows through the fifteen (15) resistor at 0 seconds?
b) At 10 seconds what voltage will have built up on the 6 micro farad capacitor?"

Thanks!

EDIT: Oh, and my attempt was to just put it in Multisim and see what happened, getting:
a) 6 Amps
b) 60 Volts

Again, don't know why because I don't understand why there's certain times in the problem...

Last edited:

#### shteii01

Joined Feb 19, 2010
4,644
I am not clear about switch J2. When does the switch J2 close?

I agree with your solution to Part A, I am also getting 6 A.

#### Supervisor

Joined Jan 17, 2009
32
Do you know HOW you got the first part?
I only see it through the err... simulation...

OH!!!
Sorry guys..
Switch two closes for those two questions.

#### shteii01

Joined Feb 19, 2010
4,644
Do you know HOW you got the first part?
I only see it through the err... simulation...

OH!!!
Sorry guys..
Switch two closes for those two questions.
Sorry about that, I should have described it.

When you are dealing with capacitor (and inductor), you have to account for the voltage across the capacitor at t=0. So. For Part A we assume that at t=0 the voltage across the capacitor is zero. That knowledge simplifies the circuit to the battery and two resistors. The two resistors are in series, you just add them up to get the equivalent resistor. Now you have battery and one resistor. Apply Ohm's Law and you get 6 A.

The others will have to double check me, but I think Part B works kinda like this:
The switch J2 is closed from t=0 to t=10, that is a long a time. The capacitor is charged. Under constant voltage, the capacitor act as an open circuit. This means that whatever voltage appears across the R2 is also appears across the capacitor C1. The voltage across R2 can be found two ways
1) Voltage divider
2) You know current through R2 from Part A. Use Ohm's Law.
Either way you solve it, it is 60 V.

#### Supervisor

Joined Jan 17, 2009
32
Sorry about that, I should have described it.

When you are dealing with capacitor (and inductor), you have to account for the voltage across the capacitor at t=0. So. For Part A we assume that at t=0 the voltage across the capacitor is zero. That knowledge simplifies the circuit to the battery and two resistors. The two resistors are in series, you just add them up to get the equivalent resistor. Now you have battery and one resistor. Apply Ohm's Law and you get 6 A.

The others will have to double check me, but I think Part B works kinda like this:
The switch J2 is closed from t=0 to t=10, that is a long a time. The capacitor is charged. Under constant voltage, the capacitor act as an open circuit. This means that whatever voltage appears across the R2 is also appears across the capacitor C1. The voltage across R2 can be found two ways
1) Voltage divider
2) You know current through R2 from Part A. Use Ohm's Law.
Either way you solve it, it is 60 V.

Wow, thanks!
I was hoping that all I had to do was just assume the capacitor was charged.
That really clears it up.

Thanks again!!!

#### hgmjr

Joined Jan 28, 2005
9,029
Hold on a moment. If the capacitor is assumed to be a short circuit when the switch t1 closes, doesn't that mean that the 150V at T0 is dropped across the 15 ohm resistor. That would then be 10 Amps rather than 6 amps. Right?

hgmjr

#### shteii01

Joined Feb 19, 2010
4,644
Hold on a moment. If the capacitor is assumed to be a short circuit when the switch t1 closes, doesn't that mean that the 150V at T0 is dropped across the 15 ohm resistor. That would then be 10 Amps rather than 6 amps. Right?

hgmjr
The resistors are in series. Resistors in series have the same current current passing through them. 6A(10)+6A(15)=150 V

#### hgmjr

Joined Jan 28, 2005
9,029
The resistors are in series. Resistors in series have the same current current passing through them. 6A(10)+6A(15)=150 V
But at t = 0 the capacitor is a short circuit across the 10 ohm resistor so no current is flowing through the 10 ohm resistor.

hgmjr

#### shteii01

Joined Feb 19, 2010
4,644
But at t = 0 the capacitor is a short circuit across the 10 ohm resistor so no current is flowing through the 10 ohm resistor.

hgmjr
Good point. I think you are right. So. We just have a 150 V battery in series with 15 Ohm resistor, which comes down to 10 A.

That changes the solution to Part B. For Part B you will have to use voltage divider to find voltage across 10 Ohm resist which will be the voltage across capacitor as well.

#### Supervisor

Joined Jan 17, 2009
32
Good point. I think you are right. So. We just have a 150 V battery in series with 15 Ohm resistor, which comes down to 10 A.

That changes the solution to Part B. For Part B you will have to use voltage divider to find voltage across 10 Ohm resist which will be the voltage across capacitor as well.
Waah!

Alright, I did some research and got why you said the capacitor acts like a short circuit and thus, the 10 ohm resistor has nothing flowing to it.

What about the next part?
Isn't the capacitor charged at this point?

EDIT: Still, according to the simulation, I see that 60 V builds up across the capacitor...
Does the 10 ohm resistor come back "into play" after 0 seconds - meaning, does the capacitor stop functioning as an open circuit after 0 seconds?

Last edited:

#### hgmjr

Joined Jan 28, 2005
9,029
I believe the answer to part B is 60V. The reason is just as you have said. 10 seconds is many times longer that 5 times RC so the cap is fully charged.

hgmjr

#### syed_husain

Joined Aug 24, 2009
61
Sorry about that, I should have described it.

When you are dealing with capacitor (and inductor), you have to account for the voltage across the capacitor at t=0. So. For Part A we assume that at t=0 the voltage across the capacitor is zero.
u can't assume the voltage accross the capacitor is zero. however u can assume unless otherwise stated that the initial stored voltage in capacitor is zero. remember, capacitor and inductor are storage elements. u r right that current is 6 A for part (a) because when the switch is closed the voltage or current can be changed abruptly in resistor but in capacitor the voltage can't be changed instantaneously. in this question, t=0, the capacitor behaves as open circuit not short circuit.

#### hgmjr

Joined Jan 28, 2005
9,029
u can't assume the voltage accross the capacitor is zero. however u can assume unless otherwise stated that the initial stored voltage in capacitor is zero. remember, capacitor and inductor are storage elements. u r right that current is 6 A for part (a) because when the switch is closed the voltage or current can be changed abruptly but in capacitor the voltage can't be changed instantaneously. in this question, t=0, the capacitor behaves as open circuit not short circuit.
I disagree with your statement that the capacitor behaves like an open circuit. At T= 0 the switch is closed and the voltage across the capacitor cannot change instantaneously. That means that at t = 0, the voltage across the 10 ohm resistor is zero and for a brief instant it presents a low impedance to ground. This briefly presents a short across the 10 ohm resistor. That means that all 150V are briefly dropped across the 15 ohm resistor. That causes 10 Amps to flow in the 15 ohm resistor initially.

hgmjr

#### syed_husain

Joined Aug 24, 2009
61
I disagree with your statement that the capacitor behaves like an open circuit. At T= 0 the switch is closed and the voltage across the capacitor cannot change instantaneously. That means that at t = 0, the voltage across the 10 ohm resistor is zero and for a brief instant it presents a low impedance to ground. This briefly presents a short across the 10 ohm resistor. That means that all 150V are briefly dropped across the 15 ohm resistor. That causes 10 Amps to flow in the 15 ohm resistor initially.

hgmjr
I = C (dV/dt) this is the relationship between capacitor current and voltage. i m now directly quoting from the book "Engineering Circuit Analysis" by Hayt, Kemmerly ( ed. 7th) from p. 217. the authors say with regarding the above equation " a constant voltage accross a capacitor results in zero current passing through it; a capacitorr is thus an "open circuit to dc" ".

#### hgmjr

Joined Jan 28, 2005
9,029
I = C (dV/dt) this is the relationship between capacitor current and voltage. i m now directly quoting from the book "Engineering Circuit Analysis" by Hayt, Kemmerly ( ed. 7th) from p. 217. the authors say with regarding the above equation " a constant voltage accross a capacitor results in zero current passing through it; a capacitorr is thus an "open circuit to dc" ".
That is an accurate statement. The only thing is that it does not apply here since at T=0 the switch is closed but instead of seeing a DC voltage, it sees an AC voltage. Somewhere in that textbook you will find a statement that says that the voltage across a capacitor cannot change instantaneously.

hgmjr

#### Supervisor

Joined Jan 17, 2009
32
u can't assume the voltage accross the capacitor is zero. however u can assume unless otherwise stated that the initial stored voltage in capacitor is zero. remember, capacitor and inductor are storage elements. u r right that current is 6 A for part (a) because when the switch is closed the voltage or current can be changed abruptly in resistor but in capacitor the voltage can't be changed instantaneously. in this question, t=0, the capacitor behaves as open circuit not short circuit.
In an effort to make sure that we get the full question (since I'm obviously leaving off some parts of the question so others don't see that I asked the Internet for help), the question fully reads as this:
When the experiment [the circuit] starts, switch one and switch three are open. Switch two is closed and both capacitors are initially discharged. At zero seconds, switch one is closed but switch three remains open.
a) What current flows through the fifteen ohm resistor at zero seconds?
b) At ten seconds, what voltage will have built up on the 6 micro farad capacitor?
c) At ten seconds, what charge will have built up on the 6 micro farad capacitor?
d) At twenty seconds, switch two is opened and switch three is closed; at thirty seconds, what will the voltage be on each of the capacitors?

Now, for my attempts:
a.) Since the capacitors are initially discharged, with switch two open the 6 micro farad capacitor acts as an open circuit (? please explain as I'm just trying to follow you guys), leaving the 15 and 10 resistor in series.
Thus: V = IR --> I = V/R --> I = 150 / (10 + 15) --> I = 6 Amps
OR
The capacitor acts as an open circuit and makes no current pass through the ten ohm resistor:
V = IR --> V/R = I ---> I = 150 / 15 --> I = 10 Amps

b.) At ten seconds, I assume that the capacitor is fully charged.
Going from what you guys are all saying, the ten ohm resistor should have the same amount of potential voltage difference as the capacitor ( again, I don't follow why this is true ) and from above, the current flowing across would be 6; multiplying that by 10 ohms, we get 60 Volts for the voltage built up across the 6 micro farad capacitor.

c.) I know [well, hope] that the equation is Q = CV
Now, if that is right, then it should be:
Q = CV --> Q = (6*10^-6)(60) = 0.00036 coulombs.
I'm using 60 Volts because I think one is supposed to use the voltage across the capacitance rather than the total.

d.) I kind of don't get how switch two can be opened with switch three closed - how do the electrons go from the capacitors back to the battery!?
Well, here's an attempt:
As you guys have been telling me, 10 seconds is a lot of time, so I assume that the capacitors are all fully charged.
Capacitors in parallel are like resistors in series - they add.
So I replaced the 6 and 3 micro farad capacitor with a 9 micro farad.
I'll just assume that when I get the voltage across the 9 micro farad capacitor that I can just multiply it by 2/3 and 1/3 to get the voltage across the 6 and 3 micro farad capacitors.
Hmm... Now what...
Is it still 60? The current would flow through the 10 ohm resistor and like in part (c), wouldn't the capacitors have a total voltage of 60 Volts?
Then 40 and 20 Volts for the 6 and 3 micro farad capacitors, respectively?

Thanks for the help so far though, guys. It doesn't discourage me but makes me want to learn more about this! ^_^

#### hgmjr

Joined Jan 28, 2005
9,029
I

Now, for my attempts:
a.) Since the capacitors are initially discharged, with switch two open the 6 micro farad capacitor acts as an open circuit (? please explain as I'm just trying to follow you guys), leaving the 15 and 10 resistor in series.
Thus: V = IR --> I = V/R --> I = 150 / (10 + 15) --> I = 6 Amps
OR
The capacitor acts as an open circuit and makes no current pass through the ten ohm resistor:
V = IR --> V/R = I ---> I = 150 / 15 --> I = 10 Amps
I will state one more time. in Part A: The current flowing in the 15 ohm resistor at the instant following the closure of the switch T1 is 10 Amps. This is true because at t = 0, the voltage across the 6uF cap is 0 volts and it is a known fact that "voltage across a capacitor cannot change instantaneously.

I am totally shocked that no other member of long standing has not jumped in to defend this answer.

hgmjr

#### syed_husain

Joined Aug 24, 2009
61
That is an accurate statement. The only thing is that it does not apply here since at T=0 the switch is closed but instead of seeing a DC voltage, it sees an AC voltage. Somewhere in that textbook you will find a statement that says that the voltage across a capacitor cannot change instantaneously.

hgmjr
at t=0 when the switch is closed, the capacitor tries to prevent the current flowing through it because voltage can't be changed instantaneously. what this means that "a sudden jump in voltage requires an infinite current and since it is physically impossible we will therfore prohibit the voltage accross a capacitor to cahnge in zero time" (Hayt, p.217). the moment switch is closed (t=0) no current will flow that means the capacitor is open circuit.

#### Supervisor

Joined Jan 17, 2009
32
I will state one more time. in Part A: The current flowing in the 15 ohm resistor at the instant following the closure of the switch T1 is 10 Amps. This is true because at t = 0, the voltage across the 6uF cap is 0 volts and it is a known fact that "voltage across a capacitor cannot change instantaneously.

I am totally shocked that no other member of long standing has not jumped in to defend this answer.

hgmjr
Alright! So, is everything only flowing through the capacitor (which is being charged) and the 15 ohm resistor because of this?
I've heard that electricity wants to flow through things that have no resistance, but just want to make sure...

How about the rest of my answers?

#### hgmjr

Joined Jan 28, 2005
9,029
at t=0 when the switch is closed, the capacitor tries to prevent the current flowing through it because voltage can't be changed instantaneously. what this means that "a sudden jump in voltage requires an infinite current and since it is physically impossible we will therfore prohibit the voltage accross a capacitor to cahnge in zero time" (Hayt, p.217). the moment switch is closed (t=0) no current will flow that means the capacitor is open circuit.
Sorry, you still have it wrong. When voltage tries to change it cannot. That means that the current actually surges. The current at the instant t=0+ increases briefly to 10 amps.

hgmjr