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 element = 10.0 V, R_1 = 43.0 k ohm, R_2 = 110 k ohm, and C = 10.5 mu F. (a) Determine the time constant before the switch is closed. s (b) Determine the time constant after the switch is closed. s (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.) I =

Explanation / Answer

After a long time with the switch open, the voltage across C will = E
By KVL, when the switch is closed, the voltage across R1 must = E and the voltage across R2 must be the voltage the cap has at t = 0 which must also = E

Thus the current thru R1 = E/R1 which must be the current out of and back into E.

The current thru R2 at t = 0 is E/R2 which is the current into and out of C. Therefore the current thru the switch = E/R1 + E/R2 *e-t/(R2*C)
After a "long time" or about 6*R2*C seconds, the cap with have discharged and the current through the switch will = E/R1 = 10/43 = 0.2325

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