Resistor value given capacitance and EMF with a resistance R after 3ms

[kalm1307]

Homework Statement

A capacitor with capacitance C=5.90 μF, initially uncharged, is connected to a source with emf ε=28.0 V and to a series resistance R. After 3.00 ms the charge on the capacitor is 100 μC. What is the value of resistor R in Ohms?

Relevant Equations

T=R/C

T=R/C
So:
0.003s=R/100μC
R=3*10^-7Ω
I am really confused with the equations I have to use

 

[berkeman]

Homework Statement:: A capacitor with capacitance C=5.90 μF, initially uncharged, is connected to a source with emf ε=28.0 V and to a series resistance R. After 3.00 ms the charge on the capacitor is 100 μC. What is the value of resistor R in Ohms?
Relevant Equations:: I am not sure

I tried with T=R/C
So:
0.003=R/100μC
R=3*10^-7
I am really confused with the ecuations I have to use

The equations you have shown are not very helpful, especially since you have not described them and have not included units.

So on a problem like this, just think in terms of the RC time constant charging equation (can you show that?), and how the voltage on the capacitor involves the charge value that the problem is asking about.

Can you please try again, post the Relevant Equations with units, and show your new attempt? Thanks!

 

[kalm1307]

The equations you have shown are not very helpful, especially since you have not described them and have not included units.

So on a problem like this, just think in terms of the RC time constant charging equation (can you show that?), and how the voltage on the capacitor involves the charge value that the problem is asking about.

Can you please try again, post the Relevant Equations with units, and show your new attempt? Thanks!

I have corrected everything you told me, do I have to post again my doubt?

 

[berkeman]

I have corrected everything you told me, do I have to post again my doubt?

I have no idea.

The relevant equation for the voltage on a charging capacitor in an RC circuit is:

$$V(t) = V_s (1- e^\frac{-t}{RC}) $$

Can you use that to figure out what the voltage is at the time specified, and translate that to the charge?

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

 

[kalm1307]

I tried this equation:
Q=C*Vb*(1-e^(-t/RC)
And I solved the equation for R as follows:

100*10^-6C=5.0*10^-6F*28V(1-e^(-0.003s/R*(5.9*10^-6F))
R=546.92Ω

 

[kalm1307]

I have no idea.

The relevant equation for the voltage on a charging capacitor in an RC circuit is:

$$V(t) = V_s (1- e^\frac{-t}{RC}) $$

Can you use that to figure out what the voltage is at the time specified, and translate that to the charge?

View attachment 300868
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

I tried this equation:
Q=C*Vb*(1-e^(-t/RC)
And I solved the equation for R as follows:

100*10^-6C=5.0*10^-6F*28V(1-e^(-0.003s/R*(5.9*10^-6F))
R=546.92Ω

 

[DaveE]

This (plus subsequent) videos:
https://www.khanacademy.org/science...se/v/ee-rc-natural-response-intuition?modal=1

 

[berkeman]

100*10^-6C=5.0*10^-6F*28V(1-e^(-0.003s/R*(5.9*10^-6F))
R=546.92Ω

I think that is close to correct, but it's extremely hard to read without LaTeX formatting, plus I think you incorrectly put C=5.0uF as the first term on the RHS instead of 5.9uF.

It is also best to solve the equations symbolically first, and then substitute in the numbers in the last step to calculate the answer. In LaTeX:
$$Q(t) = C V(t)$$
$$Q(t) = C V_s (1-e^{- \frac{t}{R C}} )$$
$$\frac{Q(t)}{C V_s} = 1-e^{- \frac{t}{R C}} $$
$$e^{- \frac{t}{R C}} = 1- \frac{Q(t)}{C V_s}$$
Then how do you simplify the LHS to get rid of the exponential so that you can keep working toward finding R?

 

[berkeman]

(BTW, if you click Reply to my last post, and toggle the "[ ]" BB code setting thing above the Edit window, you can see how easy it is to post these kind of math equations with LaTeX.)

 

[DaveE]

It is also best to solve the equations symbolically first, and then substitute in the numbers in the last step to calculate the answer.

This! It doesn't just give you the answer, it tells you a bit about how the equations work out.