Capacitance

how do you work out the capacitance of a capacitor that is charging and discharging from a V-t graph and you know the resistance is 39k ohms

i think when it is discharging, the graph decreases exponentially.
and i think you use the v=v(0)e^-t/cr equation to find r lol

you can find the time constant from the graph which is equal to resistance*capacitance.

you can find the time constant by finding the time taken for the p.d. to reach about 37% of its starting value when t=0.

do you know how ti work out the capacitance

but i don't know the capacitance

what does the actual question ask and what values are already given?

you know the resistance and can find the time constant and
time constant = [resistance] * [capacitance]

so you can find the capacitance by dividing time constant by resistance.

ok so i have a voltage of 6V and a resistor of 39000V

when its discharging i do 6 x 1/e = 2.207V (is this the time constant)
when its charging i do 6 x 2/e = 4.415V(is this the time constant)

then
C=T/r which is equal to 2.207/39000= 5.66x10^-5 F (discharging)
C=T/r which is equal to 4.415/39000= 1.132x10^-4 F(charging)