You kept the switch in position 2 for a long time. Then you flipped the switch t

Question

You 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 101 , and the battery used is 5V.

a) 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 63.2%)

______ms

b.) b) Given this time constant and the resistance R=101, what is the capacitance C of the capacitor being used?

_____ uF (micro)

Time (ms) Voltage (V) 0 0.00 5.2015 1.97 10.403 3.16 15.6045 3.88 20.806 4.32 52.015 4.97 104.03 5.00

Explanation / Answer

We know that
V(t) = V0(1 - e^[-t/RC])

now, time constant , t' => time when capacitor is 63.2 % charged
Total voltage = 5V
63.2 % of charged voltage = 3.16 V

From table, time when charge is 3.16V = t' = 10.403

B) now t' = RC
or C = 10.403 / 101 = 0.103 F

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