# finding vc

A series electrical circuit for a funfair arcade machine which you are testing, features a capacitor (C)
charging via a 1 resistor (R) and a DC supply voltage (Vs). The voltage across the capacitor (Vc) may
be described by the following equation, where t represents time.
vc = vs �1 e
−𝑡𝑡
𝑅𝑅𝑅𝑅
The arcade machine produces an output based upon the year of your birth, which is input at the start
of the game, by changing the value of the DC supply voltage (Vs). For example, if you were born in
1994 then
Vs = 1 + 9 + 9 + 4 = 23 V.
Assuming that Vc is 1V after a time of 3 seconds, determine the approximate value of the capacitor.
The formula you pasted seems to have been pasted dodgy (did you paste it directly from word or onenote?)

I'm assuming you mean $V = V_{s} (1 - e^{\frac{-t}{RC}})$

When t = 3, V = 1. You know the value of R and Vs . From the above equation, you can rearrange to find the value of C or the capacitance.

You also don't seem to have shown your working so I can't tell if your answer is correct.
(edited 1 year ago)
Im struggling with the same question. The original formula is Vc = Vs (1 - e^(-t / RC)

Vc = 1
R = 1M Ohm
t = 3

I've managed to make C the subject but I'm not sure how you can insert the numbers and get a value when you are missing both Vs & C
Original post by MumsSpaghetti
Im struggling with the same question. The original formula is Vc = Vs (1 - e^(-t / RC)

Vc = 1
R = 1M Ohm
t = 3

I've managed to make C the subject but I'm not sure how you can insert the numbers and get a value when you are missing both Vs & C

So, I don't like how the question is worded. However, I think what it is trying to say is that, if you insert the date 1994, the supply voltage is 23V and it takes 3 seconds to charge to 1V, hence Vs = 23.

You are also trying to find C so you don't need the value of C itself (after all if you have it already that kind of defeats the point of finding it). You should just have C = (some function of Vs, Vc, R, t,) and it should be as simple as plugging in the values, but if the right hand side also includes C I think you have made a mistake with your rearrangement.
(edited 1 year ago)
Original post by MouldyVinegar
So, I don't like how the question is worded. However, I think what it is trying to say is that, if you insert the date 1994, the supply voltage is 23V and it takes 3 seconds to charge to 1V, hence Vs = 23.

You are also trying to find C so you don't need the value of C itself (after all if you have it already that kind of defeats the point of finding it). You should just have C = (some function of Vs, Vc, R, t,) and it should be as simple as plugging in the values, but if the right hand side also includes C I think you have made a mistake with your rearrangement.

Yes after re-reading the question i eventually took the Vs=23 value and used that. I managed to re-arrange my formula to:

C = -t / R loge (Vc/Vs) -1

I've punched the numbers in and got a value for C but after re-inserting the value back into the original formula to verify my answer it seems to be incorrect. I got C = 1.27
Original post by MumsSpaghetti
Yes after re-reading the question i eventually took the Vs=23 value and used that. I managed to re-arrange my formula to:

C = -t / R loge (Vc/Vs) -1

I've punched the numbers in and got a value for C but after re-inserting the value back into the original formula to verify my answer it seems to be incorrect. I got C = 1.27

I'm pretty sure you rearranged wrong (although that may just be the lack of brackets), but I have the correct rearrangement in the spoiler.

Spoiler

Finally, don't forget that the resistance of the resistor is in MegaOhms, not just Ohms. Capacitors are generally measured in microfarads (10^-6), so the fact that you have a value of about one farad implies a power of ten error (an aside, it is a good idea to know the general values of some things, e.g. the drift speed of electrons in a conductor is generally a few mm or cm a second, or the speed of particles in a gas is about 300-500m/s. If you end up with something like the average speed of a gas being 10km/s in your answer then you most likely fudged up the powers somewhere)
(edited 1 year ago)