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 102 fi, and the battery used is 9V. 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%) Given this time constant and the resistance R=102H, what is the capacitance C of the capacitor being used?

Answer 1

(a) When a battery is connected to a series resistor and capacitor, the initial current is high as the battery transports charge from one plate of the capacitor to the other. The charging current asymptotically approaches zero as the capacitor becomes charged up to the battery voltage. Charging the capacitor storesenergy in the electric field between the capacitor plates. The rate of charging is typically described in terms of a time constant RC.

It is given that the battery voltage (V) = 9 V. Therefore (1-1/e) of battery voltage is: 9 x (1-1/e) = 5.69 V

Time taken for the voltage to increase to this value is 11.22 ms from the table.

Time constant = 11.22 ms.

(b) Time constant can be expressed in terms of resistance R and capacitance C as:

? = RC

or, 11.22 x 10^-3 = 102 x C

or, C = 1.1 x 10^-4 F = 110 ?F