How to solve this exponential Word problem

**daunder** · March 31st 2009, 04:19 PM

A capacitor is an RC circuit discharges according to the equation
where V is the voltage, t is the time in seconds and Vmax is the initial voltage, Determine how long it will take for the capacitor to discharge 25% of its initial charge, to the nearest hundredth of a second

V(x) = Vmax e^-2t/3

**TheEmptySet** · March 31st 2009, 04:28 PM

Originally Posted by daunder

A capacitor is an RC circuit discharges according to the equation
where V is the voltage, t is the time in seconds and Vmax is the initial voltage, Determine how long it will take for the capacitor to discharge 25% of its initial charge, to the nearest hundredth of a second

V(x) = Vmax e^-2t/3

Well you are looking for a t such that

$V(t)=.75V_{max}$

So we want to solve the equation

$.75V_{max}=V_{max}e^{-\frac{2}{3}t}$

Notice that $V_{max}$ is on both sides of the equation so it reduces out to gives

$.75=e^{-\frac{2}{3}t}$

all that is left is to solve for t

**e^(i*pi)** · March 31st 2009, 04:29 PM

Originally Posted by daunder

A capacitor is an RC circuit discharges according to the equation
where V is the voltage, t is the time in seconds and Vmax is the initial voltage, Determine how long it will take for the capacitor to discharge 25% of its initial charge, to the nearest hundredth of a second

$ V(x) = V_{max} e^{\frac{(2t)}{3}} $

I'd use v(t) as it's a function of time. I assume the equation at the bottom is given to you? You want the time when 25% is gone at which point there will be (1-0.25=0.75) left.

If so you know at time t $V(t) = 0.75V_{max}$ which will lead to $V_{max}$ cancelling.
$ e^{-\frac{2t}{3}} = \frac{V(t)}{V_{max}} = 0.75$

$ -\frac{2t}{3} = ln(0.75)$

and solve for t

**daunder** · March 31st 2009, 04:47 PM

Originally Posted by e^(i*pi)

I'd use v(t) as it's a function of time. I assume the equation at the bottom is given to you? You want the time when 25% is gone at which point there will be (1-0.25=0.75) left.

If so you know at time t $V(t) = 0.75V_{max}$ which will lead to $V_{max}$ cancelling.
$ e^{-\frac{2t}{3}} = \frac{V(t)}{V_{max}} = 0.75$

$ -\frac{2t}{3} = ln(0.75)$

and solve for t

K i didn't write the question properly
It said Determine how long it will take for the capacitor to discharge to 25% of its initial charge

so, is this correct ?
0.25 = e^-2t/3

ln 0.25 = -2t/3 ln e
t = -3ln0.25/2

**e^(i*pi)** · March 31st 2009, 05:02 PM

Originally Posted by daunder

K i didn't write the question properly
It said Determine how long it will take for the capacitor to discharge to 25% of its initial charge

so, is this correct ?
0.25 = e^-2t/3

ln 0.25 = -2t/3 ln e
t = -3ln0.25/2

Yep, that's correct

**daunder** · March 31st 2009, 05:31 PM

Thanks for your help

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