00:01
In this question, we are given the left circuit.
00:03
So, and the switch is open for a long time.
00:08
And then suddenly the switch is closed.
00:10
So when the switch is closed, the circuit becomes the one at the right.
00:16
We are given that the emf of the battery is 10 volts.
00:19
R1 is 50 kilo -ooms, r2 is 100 kilo -ooms, and the capacitor, capacitance is 10 micro -farrid.
00:28
So in part a, we want to determine the time constant before the switch is closed.
00:38
So the power before is r equivalent before times c.
00:50
So this is just r1 plus r2, the total resistance.
00:55
So you can see that the battery, the r1 and r2 is series with the capacitor.
01:05
Okay, so 150 times 10 to the 3 times 10 to the minus 6.
01:16
So we get 1 .5 seconds.
01:21
Okay, so this is the answer for pi a.
01:24
And then in part b, we need to calculate the current, the time constant after the switch is closed.
01:35
So after the switch is closed the capacitor will discharge through r2 okay so after the switch is closed the capacitor will discharge through r2 okay so after the switch is closed okay capacitor discharges through r2 okay so the time constants after is r2 times c so this is equal to times 10 to the tree times 10 to the minus 6 so you get one second okay okay then in part c so let the switch be closed at t goes to zero determine the current in the switch as a function of time okay so so just look at the circuit again okay, so the switch is here...