# Thread: How to solve this exponential Word problem

1. ## How to solve this exponential Word problem

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

2. 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

3. 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

4. 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

5. 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