In the circuit of the figure below, the switch S has been open for a long time.

Question

In the circuit of the figure below, the switch S has been open for a long time. It is then suddenly closed. Take epsilon = 10.0 V, R_i = 47.0 k ohm, R_2 = 150 k ohm, and C = 10.5 mu F. (a) Determine the time constant before the switch is closed. (b) Determine the time constant after the switch is closed. (c) Let the switch be closed at t = 0. Determine the current in the switch as a function of time. (Assume I is in A and t is in s. Do not enter units in your expression. Use the following as necessary: t.)

Explanation / Answer

current in switch = current due to battery+ current due to capacitor

= 10/47000 + 10/150000 e^(-t/1.575)

= 2.12*10^-4 + 6.67*10^-5 *e^(-t/1.575)

Note: don't enter unit as stated in question, enter answer in Ampere

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