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SCurley · Jul 2, 2009, 07:03 AM

How do I mate '-t' the subject of the formula: v=Ve(superscript)-t/RC?
The formula is about the discharging of a capacitor and I need to find out how much time needs to have passed before the voltage across the capacitor falls to a specific level.
Thanks,
SCurley.

Unknown008 · Jul 2, 2009, 07:54 AM

Is that it?

$V_C=V_Oe^{-\frac{t}{RC}}$

Put ln on both sides;

$ln(V_C)=ln(V_Oe^{-\frac{t}{RC}})$

$ln(V_C)=ln(V_O) + ln(e^{-\frac{t}{RC}})$

$ln(V_C)=ln(V_O) - \frac{t}{RC}ln(e)$

Can you continue now?

Hope it helped! :)

SCurley · Jul 2, 2009, 08:08 AM

Originally Posted by Unknown008

Is that it?

$V_C=V_Oe^{\frac{-t}{RC}}$

Put ln on both sides;

$ln(V_C)=ln(V_Oe^{\frac{-t}{RC}})$

$ln(V_C)=ln(V_O) + ln(e^{\frac{-t}{RC}})$

$ln(V_C)=ln(V_O) + \frac{-t}{RC}ln(e)$

Can you continue now?

Hope it helped! :)

Thanks for the help, but I was expecting an answer starting with -t =, or a lesson in how to transpose a formula that has a fraction to the power in it.
Could you simplify your answer, maybe that will help as it looks somewhat baffling to me.
Thanks,
SCurley

Unknown008 · Jul 2, 2009, 08:28 AM

These are already very simple (note that we are now supposed to give you the answers directly);

$ln(V_C) = ln(V_O)-{\frac{t}{RC}}ln(e)$

$ln(V_C) - ln(V_O) = -{\frac{t}{RC}}ln(e)$

Since ln (e) = 1;

$ln(V_C) - ln(V_O) = -{\frac{t}{RC}}$

Multiply both sides by RC;

$RC(ln(V_C) - ln(V_O)) = -t$

SCurley · Jul 2, 2009, 08:34 AM

Originally Posted by Unknown008

These are already very simple (note that we are now supposed to give you the answers directly);

$ln(V_C) = ln(V_O)-{\frac{t}{RC}}ln(e)$

$ln(V_C) - ln(V_O) = -{\frac{t}{RC}}ln(e)$

Since ln (e) = 1;

$ln(V_C) - ln(V_O) = -{\frac{t}{RC}}$

Multiply both sides by RC;

$RC(ln(V_C) - ln(V_O)) = -t$

OK. Thank you.
SCurley.

ali1986 · Jul 5, 2009, 04:57 PM

k=3n+2
n+1 solve for n

Unknown008 · Jul 6, 2009, 10:18 AM

Hi ali1986! Welcome to AMHD! :)

You should have created a thread of your own. Next time, do it, OK?

For your question, you cannot solve for n, but make n the subject of formula;

$k = \frac{3n+2}{n+1}$

$k(n+1) = 3n+2$

$kn+k = 3n+2$

$kn-3n = 2-k$

$n(k-3) = 2-k$

$n = \frac{k-3}{2-k}$

Hope that's what you were looking for! :)