Time for Capacitor A to Fall 10% After Discharge

[luigihs]

A 22 nF capacitor is initially charged to 10V. It is then discharged by connecting a 100k resistor across it. Approximately how long does it take for the voltage on the capacitor to fall by 10%?



v = v₀e^(–t/τ)
RC = τ = time constant



Im not sure if this the right equation what do you think guys?

 

[gneill]

The equation looks fine. How you make use of it is what's important

 

[luigihs]

τ = 22nF x 100k = 2200µs = 2.2 ms

then
v/v₀ = 0.9 = e^(–t/2.2m)
–t/2.2m = ln 0.9 = –0.1054
t = 0.0002318 = 232 µs <-- answer!

is that alright?

 

[gneill]

Yes, that's fine.

 

[rezeew]

Yes, that is the correct equation for calculating the time constant (τ) of a discharging capacitor. In this case, the capacitor has a capacitance of 22 nF and is discharged through a 100kΩ resistor. Plugging these values into the equation, we get a time constant of 2.2 ms. This means that after 2.2 ms, the voltage on the capacitor will have decreased by approximately 63.2% of its initial value (10V). To calculate the time it takes for the voltage to drop by 10%, we can use the formula t = -ln(0.9) * τ, which gives us a value of approximately 0.02 ms or 20 microseconds. This means that it takes approximately 20 microseconds for the voltage on the capacitor to fall by 10% after discharging.