The circuit below is used to determine the capacitance of an unknown capacitor b

Question

The circuit below is used to determine the capacitance of an unknown capacitor by measuring its RC constant using a known resistance R = 20.0 k omega. The capacitor is charged by flipping the switch to position 1 and waiting a long time. Then it is flipped to position 2 and the voltage across the capacitor V_ cap is measured as a function of time. Data are collected, and a line fit to In [V_ cap (t)] versus time t gives the following equation: In [V_ cap (t)] = 1.386 - 0.267 t What is the RC time constant tau? (Include units.) tau = What is the half-life T_1/2 of the exponential decay? (Include units.) T_1/2 = What is the value of the capacitance C? (Include units.) C =

Explanation / Answer

Resistance R = 20000 ohm

ln[V cap(t)] = 1.386 -0.267 t              -----------( 1)

We know V cap (t) = V bat e -t/RC

ln(V cap (t)) = ln(V bat )+ (-t/RC)        -----------( 2)

Compare (1) and(2) you get ,

t/RC = 0.267t

From this time constant RC = 1/0.267

                                         = 3.745 s

1.386 = ln(V bat )

V bat = e 1.386

         = 4 volt

(b). V cap (t) = V bat e -t/RC

    V bat / 2 = V bat e -t/RC

e -t/RC = 1/2

   -t/RC = ln(1/2)

           = -0.6931

        t = 0.6931 RC

          = 0.6931(3.745)

          = 2.595 s

(c).Capacitance C = RC/R

                           = 3.745 / 20000

                           = 1.8725x10 -4 F

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