2:32 PM session.masteringphysics.com .11 Verizon Problem 23.7 MasteringPhysics T

Question

2:32 PM session.masteringphysics.com .11 Verizon Problem 23.7 MasteringPhysics The capacitor in an RC circuit with a time constant of 13 ms is charged to 10 V. The capacitor begins to discharge at Part A At what time will the charge on the capacitor be reduced to half its initial value? Express your answer to two significant figures and include the appropriate units. t 901 us Submit Request Answer Incorrect; Try Again; 12 attempts remaining Part B At what time will the energy stored in the capacitor be reduced to half its initial value? Express your answer to two significant figures and include the appropriate units. tValue Units

Explanation / Answer

Time constant, = RC

                          = 13 ms

                          = 13 x 10^-3 s

Voltage, V = 10 V

Solution:

(A)

Given that, Q = Qo / 2

         Q = Qo e^ (- t / RC )

       1/2 = e^( - t / RC )

ln (1/2) = -t / RC

          t = - ln (1/2) * 13 x 10^-3

           = 9.01 x 10^-3 s

Ans:

Time = 9.01 ms (or) 9.01 x 10^-3 s

(B)

Given that E = Eo / 2

But 'E' is directly proportional to the square of charge 'Q'

So, E/Eo = (Q/Qo)^2

      Q/Qo = (E/Eo)

                = (1/2)

           Q = Qo e^ (- t / RC )

    (1/2) = e^( - t / RC )

ln (1/2) = -t / RC

            t = - ln (1/2) * 13 x 10^-3

             = 4.51 x 10^-3 s

Ans:

Time = 4.51 ms (or) 4.51 x 10^-3 s

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