Question

Read Activity 2-1 and Activity 2-2 in your lab manual. You have set up a circuit like that pictured in Figure 4.5, and kept the switch in position 2 for a long time. Then you flipped the switch to position 1 and measured the voltage across the capacitor as a function of time (the magnitude of the voltage is recorded below, don't worry about the sign). The resistance R has been previously measured as 104 Q, and the battery used is exactly 6V. Given the data below, what is the time constant of this exponential function (where the voltage across the capacitor is equal to 1-1/e of the battery voltage, or about 63%) Given this time constant and the resistance R, what is the capacitance C of the capacitor being used?

Answer 1

while charging a capacitor the potential increase is given as

v = e*(1-e^-t/T)

T = time constant

1)

2.36 = 6*(1-e^-(4.16*10^-3/T1))

T1 = 8.32 ms

2)

3.79 = 6*(1-e^-(8.32*10^-3/T2))

T2 = 8.33 ms

3)

4.66 = 6*(1-e^-(12.48*10^-3/T3))

T3 = 8.32 ms

4)

5.19 = 6*(1-e^-(16.64*10^-3/T4))

T4 = 8.3 ms

average time constant = T = 8.32 ms   <<------answer

(b)

T = RC

C = T/R = (8.32*10^-3)/104 = 80 uF <<------answer