# rearrange a formula to find t

#### leejohnson222

Joined Jan 11, 2023
28
so i have just started this subject and i want to get to grips understanding it for myself so it clicks.
if i need to find t from Vs = 5 sin (2pi ft - pi/3) sorry i dont have the equation written as it should be

so far i have Vs/5 - sin (2pi ft - Pi/3)
move sin to the left and becomes a negative - asin (vs/5) = 2pi ft - pi/3
asin (vs/5) + pi/3 = 2pi ft
now i can start to find t ? is this ok so far?

#### BobTPH

Joined Jun 5, 2013
6,314
Divide both sides by 2πf.

#### leejohnson222

Joined Jan 11, 2023
28
asin (Vs/5) + Pi/3
___________________
2Pi f

ok so if F= 1Mhz Vs = 3

t= asin (0.6) + 1.047
2Pi x 10^6

sorry for the format not being great, new to using maths symbols in forums

#### leejohnson222

Joined Jan 11, 2023
28
I am trying to get to grips with rearranging this formula to find c
Vc = Vs ( 1 -e -t/rc )
i can plug the numbers in to a point but im unsure about rearranging to create the new formula

Vc = 1. Vs = 22 R = 1 t= 3
at the moment i have this....
1 = 22 ( 1 - e -t/rc)
1/22 = 1 - e t/rc
0.0454 = 1 -e -3/1xc
e-3/c = 1 - 0.0454

now im struggling as i think i need to use Log here and try to break this down to a number for C

#### MrChips

Joined Oct 2, 2009
28,153
$$y = e^x$$
$$log(y) = x$$

#### leejohnson222

Joined Jan 11, 2023
28
i am quite new to this subject, could you explain a little more?

#### MrChips

Joined Oct 2, 2009
28,153
i am quite new to this subject, could you explain a little more?
log(1 - 0.0454) = -3/C

#### leejohnson222

Joined Jan 11, 2023
28
appologies i have asked two questions in the same thread, i thought i had created a new thread, however i am still trying to work both of these through

#### leejohnson222

Joined Jan 11, 2023
28
log(1 - 0.0454) = -3/C
ok so i had written down -3/c = In(0.667)
but is there not another side to this one too? or was that already cancelled out? i think i am getting this tangled

#### WBahn

Joined Mar 31, 2012
28,186
I am trying to get to grips with rearranging this formula to find c
Vc = Vs ( 1 -e -t/rc )
i can plug the numbers in to a point but im unsure about rearranging to create the new formula

Vc = 1. Vs = 22 R = 1 t= 3
at the moment i have this....
1 = 22 ( 1 - e -t/rc)
1/22 = 1 - e t/rc
0.0454 = 1 -e -3/1xc
e-3/c = 1 - 0.0454

now im struggling as i think i need to use Log here and try to break this down to a number for C
Seems your problem is the math/algebra. So let me show you how to do this properly.

Learn to track your units properly, that will help guide you along the way and help you catch mistakes as you make them.

Vc = Vs ( 1 -e -t/rc )

Also, you need to be careful about order of operations. As you have written it, you are really saying

Vc = Vs ( 1 -e -(t/r)*c )

Because division and multiplication have the same priority and are left-associative (i.e., done left to right).

Then you are saying that you are subtracting something from the value e, when you intended it to be an exponent.

You need to write it as

Vc = Vs ( 1 - e^(-t/(rc)) )

You values need units -- all physical quantities need the proper units. Without them, it's like me telling you my weight is 100. Is that really light, or pretty heavy? You have no idea, because I haven't given you a weight, I've just given you a number. A weight has to have units, such as 100 lb or 100 kg, in order to have any meaning.

So, having to guess (and engineering is NOT about guessing), I'm going to assume you mean

Vc = 1 V. Vs = 22 V, R = 1 Ω, t= 3 s

Next, work things symbolically as much as possible.

Vc = Vs ( 1 - e^(-t/(rc)) )

Vc/Vs = 1 - e^(-t/(rc))

e^(-t/(rc)) = 1 - (Vc/Vs)

-t/(rc) = ln(1 - (Vc/Vs))

-t = (rc)·ln(1 - (Vc/Vs))

c = -t / (r · ln(1 - (Vc/Vs))

NOW plug in numbers.

c = -(3 s) / (1 Ω · ln(1 - (1 V / 22 V))

c = -(3 s) / (1 Ω · ln(1 - (1 V / 22 V))

In order to take the natural log, the argument must be dimensionless, which it is since the V/V cancel out.

c = -(3 s) / (1 Ω · ln(1 - 0.0454)
c = -(3 s) / (1 Ω · ln(0.9545)
c = -(3 s) / (1 Ω · ln(0.9545)
c = -(3 s) / (1 Ω · -0.04652)
c = 64.49 s/Ω

From the equation for capacitance, we have

Q = CV

So the unit of capacitance, the farad, has units of coulombs/volt.

The unit of resistance, the ohm, has units of volts/ampere.

The one ampere is one coulomb/second.

Combining these, we get that

1 F = 1 C/V = 1 A·s/V = 1 s/(V/A) = 1 s/Ω

So

c = 64.5 F

The next thing you should do is verify the correctness of the answer by plugging it into the original problem.

Vc = Vs ( 1 - e^(-t/(rc)) )

Vc = (22 V) · ( 1 - e^(-3 s/(1 Ω · 64.5 F)) )

1 Ω·F = 1 s (use the relations above to confirm this.

Vc = (22 V) · ( 1 - e^(-3 s / 64.5 s) )
Vc = (22 V) · ( 1 - e^(-3/64.5) )
Vc = (22 V) · ( 1 - e^(-3/64.5) )
Vc = (22 V) · ( 1 - e^-0.04651 )
Vc = (22 V) · ( 1 - e^-0.04651 )
Vc = (22 V) · ( 1 - 0.9546 )
Vc = (22 V) · ( 1 - 0.9546 )
Vc = (22 V) · (-0.04544)
Vc = (22 V) · (-0.04544)
Vc =0.9998 V
Vc = 1.000 V (check)

If you have any questions about any step in the above work, ask. We can then focus on specific stumbling blocks.

#### leejohnson222

Joined Jan 11, 2023
28
this is extremely helpful and i like the confirmation of working the formula again to confirm your correct
I have been reading and watching lectures all day so i need a break, but will come back to this as i have questions
since i think i lost my way when i needed to look at working on the natural log stage and breaking that down
for ref this is how my formula actually looks (attached)

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#### leejohnson222

Joined Jan 11, 2023
28
i am following the formula transposition, how ever when i get to putting the numbers in
at this point
c = -(3 s) / (1 Ω · ln(0.9545)
i miss one of your steps and arrive directly at the answer 64.4 F

#### WBahn

Joined Mar 31, 2012
28,186
i am following the formula transposition, how ever when i get to putting the numbers in
at this point
c = -(3 s) / (1 Ω · ln(0.9545)
i miss one of your steps and arrive directly at the answer 64.4 F
I'm not sure what you're asking. If you evaluate that expression you get 64.4 F. What step do you think your're missing?

If you go step by step you get 64.4 s/Ω.

But 1 s/Ω is 1 F.

You can either work through the fundamental definitions of the various units to show that, like I did above, or use some relationships that are (or will become) easy to remember.

We work all the time with RC time constants, so we know that units of resistance multiplied by units of capacitance yield units of time. In particular

1 Ω · 1 F = 1 s

So

1 F = 1 s / 1 Ω = 1 s/Ω

#### leejohnson222

Joined Jan 11, 2023
28
its fine its my mistake, i ran through it a few times and came to same answer, probably just tired of looking at my computer and reading

#### dl324

Joined Mar 30, 2015
15,511
for ref this is how my formula actually looks (attached)
Formulas will be easier to read if you use LaTeX:
$$V_C=V_S(1-e^\frac{-t}{RC})$$
$$\frac{V_C}{V_S}=1-e^\frac{-t}{RC}$$
$$e^\frac{-t}{RC}=1-\frac{V_C}{V_S}$$
$$\frac{-t}{RC}=ln(1-\frac{V_C}{V_S})$$
$$t=-RCln(1-\frac{V_C}{V_S})$$
I showed more intermediate steps than I'd normally use to make it easier to follow the math.