2132016 1155 PM 711100 Gradebook Print Calculator Periadic T

2/13/2016 11:55 PM 71.1/100 Gradebook Print Calculator Periadic Table Question 12 of 14 Map sapling learning 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? 4.5 F 80 9 V Number 60 20 2.0 HF Assuming the switch remains closed for a long time, which capacitor will be the first to reach 95% of its final charge level? O The 4.5-uF capacitor O The 2.0-uF capacitor What is the time constant for charging this capacitor? Number Previous Give Up & View Solution O Check Answer Next Exit Hint

Solution

Initially a fully discharged capcitor acts as an short circuit and drawns maximum current across its ends.

So Now

1/ R = 1/ 80 + 1/20 + 1/60 = ( 3 + 12 + 4 )/ 240 = 19/240

R = 12.63 Ohms

3) Time constanct is

I = V/ R = 9/ 12.63 = 0.712 amperes

2 ) The equation we have to use now is

= Q / Qmax = 1- e^(-t/RC)

For capcitor with 2.0 micro farads

= 0.95 = 1 - e^(-t/ 80x2x10-6)

=>t =4.79 x 10^-4 seconds

For capacitor with 4.5 micro farads

=0.95 = 1- e^(-t/ 20X 4.5 x10^-6)

=> t =2.69 x 10^-4 seconds

Therefore answer is 4.5 micro farad capacitor.

3) Time constant is 20x4.5 = 90 micro seconds.