# Capacitor and voltage source in one?

Gold Member

## Homework Statement

Consider circuit below.

(a)Use integral calculus to determine how much heat is generated through R when switch S is closed.

(b)Derive the expressions for the charge q(t) and the current i(t) as functions of time, for the discharging capacitor in the circuit.

http://img516.imageshack.us/img516/134/picture5fy6.th.jpg [Broken]

## The Attempt at a Solution

First of all I'm not sure what the capacitor is or the voltage source is. Is it a battery or a capacitor??
It seems that it is one and both, but how can this be?

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Hootenanny
Staff Emeritus
Gold Member
The voltage source in the circuit is the capacitor. All a voltage source does is provide a potential difference across the circuit, in this case the potential difference is provided by the potential across the plates of the capacitor.

Gold Member
The voltage source in the circuit is the capacitor. All a voltage source does is provide a potential difference across the circuit, in this case the potential difference is provided by the potential across the plates of the capacitor.

for part a where I have to

(a)Use integral calculus to determine how much heat is generated through R when switch S is closed.

I know
$$\Delta Vo= 100V$$
$$C= 20x10^{-6} F$$
$$R= ?$$

and that $$E_{int}= Q$$ in a resistor (heat produced), and thus the heat that is generated is in the form of the internal energy that builds up in the resistor.
And also that the rate at which the energy is converted to heat in the resistor is
$$P= I \Delta V$$

but as to the equation I'm not sure how to set it up.

I also don't know R or is it not necessary to know R?

Thanks

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Here's what happened:
You connected your capacitor to a voltage source with 100 V ..
After long time, when no current flows, the voltage across cap is 100 V
(This is step response)

and now, you take capacitor and connect to a resistor (second thevenian circuit)
you u know
i = C.dv/dt
so, i does not change intially, but v makes a sudden jump
use x(t) = .. equation
(see Natural response section in your book .. you will see detailed steps)

and you don't need to know R,
where does the energy lost from the capacitor is going?
Ans: to R
and using conservation of energy

Gold Member
Here's what happened:
You connected your capacitor to a voltage source with 100 V ..
After long time, when no current flows, the voltage across cap is 100 V
(This is step response)
I get that. and it's because there is no current flow at the time when the capacitor is fully charged.
and now, you take capacitor and connect to a resistor (second thevenian circuit)
I don't get this. Isn't there only 1 switch there? so how do I take the capacitor and connect it to the resistor if the capacitor is already connected to the resistor since it produces the potential difference in the first place?

you u know
i = C.dv/dt
yes
so, i does not change intially, but v makes a sudden jump
how does v make a sudden jump if the current is decreasing? and the V is increasing at the same rate? (I think that's what happens when you charge a capcitor)
use x(t) = .. equation
(see Natural response section in your book .. you will see detailed steps)
I can't find that, it would be in which section exactly?(under what topic..under electricity and circuits)
and you don't need to know R,
where does the energy lost from the capacitor is going?
Ans: to R
and using conservation of energy

Thanks.

how does v make a sudden jump if the current is decreasing? and the V is increasing at the same rate? (I think that's what happens when you charge a capcitor)

Thanks.

ook, say makes a negative jump I can't find that, it would be in which section exactly?(under what topic..under electricity and circuits)
If you are electrical, and using electric circuits book then it shouldn't be hard

See page 2 (Transients) for quick solution:
http://www.ece.uwaterloo.ca/~ece100/Final_W08_formulas.pdf [Broken]

OR

and see my prof notes (for finding x(t) ... ) for little 30 seconds explanation:
http://www.ece.uwaterloo.ca/~ece100/
---> http://www.ece.uwaterloo.ca/~ece100/Canizares_notes/
---> http://www.ece.uwaterloo.ca/~ece100/Canizares_notes/SummaryRL_RC_AC.pdf
(this is what you need: a 5 minutes read)

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Gold Member
ook, say makes a negative jump If you are electrical, and using electric circuits book then it shouldn't be hard
I'm using a physics book.
See page 2 (Transients) for quick solution:
http://www.ece.uwaterloo.ca/~ece100/Final_W08_formulas.pdf [Broken]

OR

and see my prof notes (for finding x(t) ... ) for little 30 seconds explanation:
http://www.ece.uwaterloo.ca/~ece100/
---> http://www.ece.uwaterloo.ca/~ece100/Canizares_notes/
---> http://www.ece.uwaterloo.ca/~ece100/Canizares_notes/SummaryRL_RC_AC.pdf
(this is what you need: a 5 minutes read)

I'm even more confused now. I read the 2nd link and just got through 1/2.
It helped but I got up to power and then capacitor later. Then the notes mention energy in a capacitor as being $$W_{c}(t)= 1/2 CV^2(t)= 1/2 \frac{q^2(t)} {c}$$ (pg 77 second link)and since you said that all the energy goes from the capacitor to the resistor does this happen with this situation and thus the energy can be calculated from the above equation??

and I don't know what the notation of x(t) is supposed to be anyhow. even though I read the derivation which wasn't really a derivation in the first place.

BUT I do have a derivation in my book of the enrgy that is delivered to a resistor during the time interval it is discharged after it is charged to full potential difference of a battery.
The equation that is solved is $$E_r= 1/2 C \epsilon^2$$

is this the same as what I'm asked above, where I have to find the heat but the heat is from the energy that the resistor gets from the capacitor?.

Thanks.

Note: if anyone would like to help me out on this problem it'd be appreciated.

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Hootenanny
Staff Emeritus
Gold Member
BUT I do have a derivation in my book of the enrgy that is delivered to a resistor during the time interval it is discharged after it is charged to full potential difference of a battery.
The equation that is solved is $$E_r= 1/2 C \epsilon^2$$

is this the same as what I'm asked above, where I have to find the heat but the heat is from the energy that the resistor gets from the capacitor?.
You are correct, all the energy stored in the electric field by the capacitor will be dissipated by the resistor as heat.

Gold Member
You are correct, all the energy stored in the electric field by the capacitor will be dissipated by the resistor as heat.

so would the ammount of heat be

$$E_r= 1/2 C \epsilon^2 = 1/2(20x10^{-6})(100V)^2= 0.1J$$ ?

Hootenanny
Staff Emeritus
$$E_r= 1/2 C \epsilon^2 = 1/2(20x10^{-6})(100V)^2= 0.1J$$ ?
Looks good to me 