# CapacitanceWatch

Thread starter 1 year ago
#1
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
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1 year ago
#2
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
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1 year ago
#3
(Original post by man111111)
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
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.
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Thread starter 1 year ago
#4
(Original post by btsseokjin)
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
do you know how ti work out the capacitance
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Thread starter 1 year ago
#5
(Original post by akaash13)
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.
but i don't know the capacitance
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1 year ago
#6
(Original post by man111111)
do you know how ti work out the capacitance
what does the actual question ask and what values are already given?
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1 year ago
#7
(Original post by man111111)
but i don't know the capacitance
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.
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Thread starter 1 year ago
#8
(Original post by btsseokjin)
what does the actual question ask and what values are already given?

ok so i have a voltage of 6V and a resistor of 39000V (i need to find the capacitance)

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)
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Thread starter 1 year ago
#9
(Original post by akaash13)
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)
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1 year ago
#10
(Original post by man111111)
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)
The time constant Is the time taken to reach that voltage from when you start discharging, not the voltage itself. I.e. It would be the time taken discharging from 6v to 2.207v which you can read from the graph
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