Please help with the attached homework problem Consider the

Please help with the attached homework problem!

Consider the circuit shown in the diagram below. Before the switch is closed, both capacitors are uncharged. Immediately after the switch is closed, what is the amount of current supplied by the battery? Assuming the switch remains closed for a very long time, which capacitor will be the first to reach 95% of its final charge level? What is the time constant for charging this capacitor?

Solution

It needs to understood here that as the current is switched on for a capacitor it acts like a closed circuit.

That is, we only need to concern ourselves with the net resistance of the circuit for the moment the switch is turned on.

a.) 1/ Rnet = 1/ 80 + 1/20 + 1/70 = 7+28+8/560 = 43 / 560

or Rnet = 13.0233 Ohms

That is the current supplied by the battery = V/R = 6 / 13.0233 = 0.4607 A

b.) Now, once the current is turned on, each of the resistors will start getting charged. For any capacitor, the charge is given as Q = Qs(1 - e^-t/RC)

or, 0.95 = 1- e^-t/RC

or e^-t/RC = 0.05

or t / RC = 2.9957

or t = 2.9957 x RC

Hence, the 4.5 uF capacitor will be the first to get charged to 95 % of its final charge. And the time constant for this would RC = 9 x 10^-6 seconds