Question 26 of 28 Incorrect Mapoob sapling learning An initially uncharged 4.81

Question

Question 26 of 28 Incorrect Mapoob sapling learning An initially uncharged 4.81 × 10° F capacitor and a 8210 resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit? Calculate the circuit's time constant. How much time must elapse from the closing of the circuit for the current to decrease to 2.43% of its initial value? Initial current: Number 0.0001827A Time constant: Number 0.03949 Elapsed time: Number 1.17x 17s Previous O Check Answer 0 Next Exit Hint

Explanation / Answer

given

capacitance of capacitor C = 4.81*10^-6 F, resistor R = 8210 ohms, potential difference by battery V = 1.50 V
we know that

  
   in RC circuit as soon as the battery is connected in the circuit there would be large potential difference between the terminals of the capacitor (while charging)

   so the more number of electrons can cross the cross sectional area per unit time , later it will be decreases gradually become zero if the capacitor is fully charged.

will be given as I(t) = I_0 e^-(t/T)

                   T is time constant of RC circuit

now initila current in the circuit is I = V/R   = 1.5/8210 = 0.0001827 A

time constant is T = RC = 8210*4.81*10^-6 s = 0.0394901 s

elapsed time for 2.43 % current is I(t) = Ie^-(t/T)

       given I(t) = 0.0243I, I = 0.0001827 A, T = 0.0394901 s

   substituting the above data

               0.0243 I = Ie^(-t/0.0394901)

               t = 0.1468s

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