Calculating Time to Discharge Capacitor to 3V

[Rupturez]

Homework Statement

Determine how long it takes the capacitor to discharge to a value of 3v
C=1.5E-6F
R=1E3ohms
Vi=25v
Vf=3v

Homework Equations

T=RC
v=Vi*e^(t/T)

The Attempt at a Solution

T=1000*1.5E-6
T=1.5E-3s

Vf(3v) is 12% of the initial 25v

Im having trouble solving for time with this equation v=Vi*e^(t/T)

many thanks in advance

 

[Rupturez]

Should I be solving for t or using a normalised universal timeconstant(RC) curve to estimate time?

 

[jambaugh]

Homework Statement

Determine how long it takes the capacitor to discharge to a value of 3v
C=1.5E-6F
R=1E3ohms
Vi=25v
Vf=3v

Homework Equations

T=RC
v=Vi*e^(t/T)

The Attempt at a Solution

T=1000*1.5E-6
T=1.5E-3s

Vf(3v) is 12% of the initial 25v

Im having trouble solving for time with this equation v=Vi*e^(t/T)

many thanks in advance

Firstly $V = V_i e^{-t/T}$ there's a negative sign reflecting the decay of the voltage... it doesn't grow!

Secondly solving this for t is basic algebra, apply inverse operations until you isolate t.
$A = B e^{t/C}$
$A/B = e^{t/C}$
$\ln(A/B) = t/C$
$C\ln(A/B) = t$

 

[Rupturez]

Thanking you jambaugh

 

[rwisz]

I would approach this problem by using the equation T=RC, where T is the time constant, R is the resistance, and C is the capacitance. In this case, T=1000*1.5E-6=1.5E-3s. This means that it will take approximately 1.5 milliseconds for the capacitor to discharge to 1/e (approximately 37%) of its initial voltage.

To calculate the time it takes for the capacitor to discharge to 3V, we can use the equation v=Vi*e^(t/T), where v is the voltage at any given time, Vi is the initial voltage, and t is the time. In this case, we know that v=3V, Vi=25V, and T=1.5E-3s. We can rearrange the equation to solve for t, giving us t=T*ln(v/Vi)=1.5E-3*ln(3/25)≈0.00047 seconds.

It's important to note that this is an ideal calculation and may not perfectly match the actual time it takes for the capacitor to discharge in real-world conditions. Factors such as internal resistance, leakage, and fluctuations in the power source can affect the actual discharge time.