Capacitance Watch

man111111
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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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btsseokjin
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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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akaash13
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(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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man111111
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(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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man111111
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(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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btsseokjin
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(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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akaash13
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(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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man111111
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(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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man111111
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(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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akaash13
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(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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