- #1

spsch

- 111

- 21

- Homework Statement
- A capacitor with Capacity C, a Battery with Voltage ## V_0 ## and Resistance R are connected.

C = 1mF, R = 500 Ohms, V_0 = 10V

a) Set up a differential equation showing the change in voltage over time.

b) draw a slope field of the equation

c) solve the equation

d) show that ## V(t) =V_0(1-e^(\frac{-t}{RC})) is a solution

e) draw the function

- Relevant Equations
- C = Q/V, The voltage difference around a closed loop is zero.

Hi all.

I have another exam question that I am not so sure about. I've solved similar problems in textbooks but I have a feeling once again that the correct way to solve this problem is much simpler and eluding me.

Especially because my answer to a) is already the solution to c) and d) (I did my way before looking at c and d) and I can't figure out b with my equation.

So I set up ## V_0 -IR -\frac{Q}{C} = 0 ##

I separated and got to ## Q = V_0C(1-e^(\frac{-t}{RC}))##

Then because Q/C = V I divided both sides by the Capacitance and got:

## V = V_0(1-e^(\frac{-t}{RC}))##

(Solution to c and d)

Now for b If I take the time derivative again I end up with ## \frac{dv}{dt} = -V_0(RC)(1-e^(\frac{-t}{RC})) ##

Which I guess I could draw but it seems to complicate to be correct?

If there is another way to start, a better way, could someone point me in that direction without giving the answer away?

Thank you all!

I have another exam question that I am not so sure about. I've solved similar problems in textbooks but I have a feeling once again that the correct way to solve this problem is much simpler and eluding me.

Especially because my answer to a) is already the solution to c) and d) (I did my way before looking at c and d) and I can't figure out b with my equation.

So I set up ## V_0 -IR -\frac{Q}{C} = 0 ##

I separated and got to ## Q = V_0C(1-e^(\frac{-t}{RC}))##

Then because Q/C = V I divided both sides by the Capacitance and got:

## V = V_0(1-e^(\frac{-t}{RC}))##

(Solution to c and d)

Now for b If I take the time derivative again I end up with ## \frac{dv}{dt} = -V_0(RC)(1-e^(\frac{-t}{RC})) ##

Which I guess I could draw but it seems to complicate to be correct?

If there is another way to start, a better way, could someone point me in that direction without giving the answer away?

Thank you all!

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