- #1

naissa

- 2

- 0

In the lab, I charged an discharged a capacitor. The voltage decreased with the equation V=Voe^(-t/RC)

Can you explain me this lab?

Thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter naissa
- Start date

- #1

naissa

- 2

- 0

In the lab, I charged an discharged a capacitor. The voltage decreased with the equation V=Voe^(-t/RC)

Can you explain me this lab?

Thanks

- #2

Curious3141

Homework Helper

- 2,858

- 88

Do you know how to set up and solve first order linear differential equations ? You'll need to know that in order to understand the derivation of that equation.

- #3

naissa

- 2

- 0

yes I do

but I do not understand the cocnepts of the circuit.

- #4

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

- #5

Jelfish

- 144

- 5

First let's draw a picture of the circuit:

Code:

```
C
+-----| |------+
| |
| |
+-----~~~------+
R
```

[tex]IR+\frac{Q}{C}=0[/tex]

Since the voltage we're interested in is the voltage across the resistor, we can substitute IR for V:[tex]V+\frac{Q}{C}=0[/tex]

[tex]V=-\frac{Q}{C}[/tex]

We want to know how these things change with time, so we take the derivative with respect to time:[tex]V=-\frac{Q}{C}[/tex]

[tex]\dot{V}=-\frac{\dot{Q}}{C}[/tex]

You should know that the current is the time-derivative of charge:[tex]\dot{V}=-\frac{I}{C}[/tex]

You also know that the current will be proportional to the voltage (Ohm's Law):[tex]\dot{V}=-\frac{V}{RC}[/tex]

So now, doing a little rewritting and rearranging, we get:[tex]\frac{dV}{V}=-\frac{dt}{RC}[/tex]

From here, you should be able to solve the differential equation. Hope that helps!- #6

BobG

Science Advisor

Homework Helper

- 330

- 84

Current will only flow through a capacitor until it reaches the same voltage as the source. If you apply a DC voltage to the RC circuit, current flows only until the capacitor charges up (you're measuring your voltage drop across the resistor?). Then the circuit reaches an equilibrium where the capacitor becomes an open circuit. Your lab is measuring how long this takes?naissa said:

In the lab, I charged an discharged a capacitor. The voltage decreased with the equation V=Voe^(-t/RC)

Can you explain me this lab?

Thanks

If so, you should be substituting in different capacitors and/or resistors and should discover the time required to charge a capacitor in an RC circuit is very predictable. Obviously, the fact that the value of the resistor and capacitor are in the equation implies that, but you're trying to find the precise relationship.

You could also find the precise relationship mathematically.

If t=0, what's the voltage?

If t=RC, what's the voltage? (this is particularly important)

How long until the voltage is practically zero; in other words, how long to charge the capacitor? (this is a little arbitrary - when t is 5 times larger than RC, the result is considered close enough to zero to be ignored).

This is important even for AC circuits (the type where capacitors and inductors are most useful), since the capacitor is going to take time to reach equilbrium with the rest of the circuit. When you first turn on your AC circuit, there will be a very short period of time where the performance of the circuit doesn't match what you want for the long term performance. That short time to charge up is called the transient response and is what you're calculating in your lab. The long term performance is called the steady state.

The others gave a good explanation of how the equation you're using was derived - in other words, why the equation does such a good job explaining what you should see in your lab.

Last edited:

- #7

Antiphon

- 1,683

- 3

The lower the voltage on the capacitor, the more slowly the current will come out.

As the current comes out of the capacitor, the voltage drops.

Imagine a water tank with a small hole in the bottom. The voltage

is like the pressure. As the water level (charge) comes down the

pressure (voltage) decreases and so the water coming out the hole

(current) slows down (or is less).

At the end you have barely a trickle when at first you had a strong

stream.

As the current comes out of the capacitor, the voltage drops.

Imagine a water tank with a small hole in the bottom. The voltage

is like the pressure. As the water level (charge) comes down the

pressure (voltage) decreases and so the water coming out the hole

(current) slows down (or is less).

At the end you have barely a trickle when at first you had a strong

stream.

Last edited:

Share:

- Last Post

- Replies
- 1

- Views
- 85

- Replies
- 1

- Views
- 333

- Replies
- 4

- Views
- 345

- Last Post

- Replies
- 12

- Views
- 632

- Replies
- 8

- Views
- 572

- Last Post

- Replies
- 1

- Views
- 227

- Replies
- 3

- Views
- 384

- Replies
- 13

- Views
- 655

- Last Post

- Replies
- 3

- Views
- 775

- Last Post

- Replies
- 1

- Views
- 207