Help - Isaac Physics - Discharging Capacitor

Hiya!
I would be grateful for any help on part d of the Discharging Capacitor Isaac Physics Question.

Question:
In the circuit below, the capacitor is initially holding a charge of Q_0. The capacitance of the capacitor is C and the resistors have values R_1 and R_2.

Part A:

Part B:

Part C:

Part D:
At the point when the capacitor is holding a charge of Q how much energy has been dissipated in R_2?

I have looked at all of the hints but don't know where to start.

(edited 3 years ago)

Reply 1

fastnfuriouseng

OP

5

Hiya!
I would be grateful for any help on part d of the Discharging Capacitor Isaac Physics Question.

Question:
In the circuit below, the capacitor is initially holding a charge of Q_0. The capacitance of the capacitor is C and the resistors have values R_1 and R_2.

Part A:

Part B:

Part C:

Part D:
At the point when the capacitor is holding a charge of Q how much energy has been dissipated in R_2?

I have looked at all of the hints but don't know where to start.

(edited 3 years ago)

Original post by fastnfuriouseng

Hiya!
I would be grateful for any help on part d of the Discharging Capacitor Isaac Physics Question.

Question:
In the circuit below, the capacitor is initially holding a charge of Q_0. The capacitance of the capacitor is C and the resistors have values R_1 and R_2.

Part A:

Part B:

Part C:

Part D:
At the point when the capacitor is holding a charge of Q how much energy has been dissipated in R_2?

I have looked at all of the hints but don't know where to start.

Don't abuse the forum by duplicating thread. Restrict the question within the original thread and don't bump the thread in less than 1 day.
Read the forum posting guidelines.

As for your question, you need the graph of Figure 10.39(d) in the following link.
https://openstax.org/books/university-physics-volume-2/pages/10-5-rc-circuits

The voltage across R2 decreases exponentially and you can use the following expression to find the energy has been dissipated in R_2:

E = \dfrac{V^2}{R_2} \Delta t

Reply 3

fastnfuriouseng

OP

5

Original post by Eimmanuel

Don't abuse the forum by duplicating thread. Restrict the question within the original thread and don't bump the thread in less than 1 day.
Read the forum posting guidelines.

As for your question, you need the graph of Figure 10.39(d) in the following link.
https://openstax.org/books/university-physics-volume-2/pages/10-5-rc-circuits

The voltage across R2 decreases exponentially and you can use the following expression to find the energy has been dissipated in R_2:

E = \dfrac{V^2}{R_2} \Delta t

Sorry! I just made a mistake in my first message and tried clearing it and thought it wouldn't pop up anymore.

Thank you for your help! However, I tried using the equation you have given and don't seem to get the correct answer...I used the previous parts and simplified to get

Could you please help me out a bit more? :smile:

Original post by fastnfuriouseng

Sorry! I just made a mistake in my first message and tried clearing it and thought it wouldn't pop up anymore.

Thank you for your help! However, I tried using the equation you have given and don't seem to get the correct answer...I used the previous parts and simplified to get

Could you please help me out a bit more? :smile:

How do you use the equation?

Spoiler

Reply 5

fastnfuriouseng

OP

5

Original post by Eimmanuel

How do you use the equation?

Spoiler

I squared the initial voltage (see part A) and divided this by R_2 and multiplied it by delta v (see part C). Could you maybe give me an extra pointer? I am stuck am not that sure how to use the formula correctly....

Original post by fastnfuriouseng

I squared the initial voltage (see part A) and divided this by R_2 and multiplied it by delta v (see part C). Could you maybe give me an extra pointer? I am stuck am not that sure how to use the formula correctly....

I did mention in post #3 that you need graph in Fig10.39(d). The graph in Fig10.39(d) allows you write an expression of voltage across R2 and you need to square this expression.
The last crucial part is using calculus in the following formula.

E = \dfrac{V^2}{R_2} \Delta t

Reply 7

happyapple5

11

This is what I've got but its not correct, I integrated delta t using the limits from 0 to (answer to part C)
my only worry is that the V^2 part is incorrect
could someone please look at this
I've looked at the link in the previous posts of this thread
and get V^2 to be the part on top of the fraction.
I've also tried -ln(Q_0/Q) (V^2 part)

(edited 3 years ago)

Reply 8

happyapple5

11

I have had another go at this question and this is my follow up answer. (after learning some log laws)
I still have no clue what I'm doing wrong
I've used calculus to get this answer by using the answer to part c
discharge .png

here are my workings

(edited 3 years ago)

Original post by happyapple5

I have had another go at this question and this is my follow up answer. (after learning some log laws)
I still have no clue what I'm doing wrong
I've used calculus to get this answer by using the answer to part c
discharge .png

here are my workings

What is the function that you are integrating? It is not shown in your working.
If you think is a constant function, see Fig 10.40(d) in the following link.
https://openstax.org/books/university-physics-volume-2/pages/10-5-rc-circuits

In a capacitor, the discharge function cannot be a constant function, it does not make sense.
Hopefully, you can relate maths to the physical situation.

Reply 10

AshRevise

11

Find total energy released by capacitor up to point Q (using your answer to part A may be helpful).

The energy released by a resistor is inversely proportional to its resistance. Therefore take this answer and multiply it by R_1/(R_2+R_1).

The answer I got worked on Isaac Physics...not sure if it's the best solution but it works!

Help - Isaac Physics - Discharging Capacitor

Related discussions

Latest

Trending

Trending

Articles for you