# How long does it take to reach 50% of V1?

1. Jun 10, 2015

### Matt Howard

I'm unsure of the best way to go about finding out the time it takes for the capacitor to reach a voltage of 50% of V1.

V1: 14V
Resistor before 2pt switch: 1K ohm
Resistor after 2pt switch: 220 ohm
(Charging) 1 time constant is 1.7mS
(Discharging) 1 time constant is 308 micro seconds

I've attached a picture of the circuit with the values labelled.

Is there an equation for working this out or is there a specific way of working it out? Any help or suggestions are greatly appreciated - I'm not looking for anyone to do the work for me just looking for a jab in the right direction

I thought I could work it out based on the idea that at 1TC the capacitor has reached 63% of the supply voltage (8.82V) and working out the difference there but kept getting myself in a right muddle.

2. Jun 10, 2015

### Svein

You know that charging a capacitor goes according to $V=V_{s}(1-e^{\frac{-t}{T}})$? And discharging goes as $V=V_{0}(e^{\frac{-t}{T}})$?

3. Jun 10, 2015

### Matt Howard

I think I vaguely remember that from class. so for me it would be something like: 7=14(1-e-t/1.7x10-3)? Is that right?
I'm assuming it's t I'm looking for so how would I go about transposing that to solve for t?

4. Jun 10, 2015

### SteamKing

Staff Emeritus
What have you tried?

There's some algebraic simplifications which can be done the equation before solving for t.

5. Jun 10, 2015

### Matt Howard

So simplified is 7=14-14e-t/1.7x10-3

The only idea I could find (Until Svein commented) was working it out from the fact that the capacitor is 63% charged when 1 time constant had lapsed. I worked out the voltage from that and thought I could work out the time from the differences in percentage and voltage. I.e a very confusing way of doing it that a friend suggested.

6. Jun 10, 2015

### psparky

Yep, that is correct.

Solve for t.

ln .5 = -t/(1.7E-3)

t = 1.18 ms.

7. Jun 10, 2015

### psparky

This sounds like nonsense. Setting the capacitor voltage to 7 volts in the formula is good. What is said above probably isn't going to do you any favors.
You don't care when it's charged, you just care when it's 7 volts.

8. Jun 11, 2015

### Staff: Mentor

The assumption being made is that the switch is moved from its lower position (shown) to its upper position at t=0. Is that what you're told?

I ask only because a different switching scenario is possible.

9. Jun 11, 2015

### Matt Howard

That is correct, the switch is being moved from the lower position to the upper position at t=0

10. Jun 11, 2015

### Staff: Mentor

So .... you're all sorted now?

The unfamiliar step was that where you need to take natural logarithms of both sides, in the process losing the exponential function.

11. Jun 11, 2015

### Matt Howard

Yes, I am sorted now, thank you very much! I have now completed the work, thanks again!