Question

Use your graphs to measure the time required for the voltage across the capacitor to achieve 63% of the total change in voltage. That is, for charging, the time needed to rise from 0 volts to 63% of the final voltage. (Note your final voltage may not be exactly 5 V.) For discharging, it is the time required to fall from the initial voltage to a value just 37% of the initial value. Compare these times with the calculated time constants. Calculate the maximum amount of charge stored on the capacitor. (Hint: What equation can you use to relate the maximum potential difference across the capacitor to the maximum charge stored on it?) Make sure your answer for the maximum charge includes your uncertainty in your answer.

Answer 1

`t = \tau`

`Q_(max) = CV_o`

Step-by-step explanation:

during charging of the capacitor we know that the voltage across the capacitor is given as

`V = V_o(1 - e^(-t/\tau))`

now the time to raise the voltage across the capacitor is given from 0 to 63%

so it is

`V = 0.63 V_o`

`0.63 V_o = V_o(1 - e^(-t/\tau))`

`1 - e^(-t/\tau) = 0.63`

`0.37 = e^(-t/\tau)`

`t/\tau = 1`

`t = \tau`

Similarly we know that during discharging we have

`V = V_o e^(-t/\tau)`

so here this will decrease by 37 %

`0.37 V_o = V_o e^(-t/\tau)`

again we have

`t = \tau`

Also we know that during charging the maximum charge on the capacitor is

`Q = CV`

`Q = CV_o(1 - e^(-t/\tau))`

for maximum value of charge t = infinite

so we will have

`Q_(max) = CV_o`