P5) (20 pts) RC Circuit: Consider the circuit shown below....

P5) (20 pts) RC Circuit: Consider the circuit shown below. The capacitor is initially uncharged. (a) If the switch, S, flipped to the left at time t=0 and then you find that the voltage across the capacitor is 7.00 V after 2.00 ms, find the time constant, ????C = RC. (b) Use your value of the time constant to determine the capacitance of the capacitor in the RC circuit. (If you absolutely cannot get part (a), ask me for a value you can use in this step.) (c) Now, if you wait a very, very, very, very long time, (we are talking hours here!) how much charge will be on the capacitor? (d) Now, if the switch is flipped to the right, what is the time constant of the discharging circuit? (SHOW ALL WORK) Text or e-mail me if you don't know how to get started on this one. 10.0 V 25.0 k????? #1 #2 75.0 k????? C

Transcript

00:01 When the switch is moved to left, it is in position 1.

00:07 The capacitor will charge through a 25 kilo -on resistance and using a 10 -volt battery.

00:17 Right.

00:18 So we know when the capacitor is charged through a resistance and a battery, at any moment of time, the potential of the capacitor v is equal to e is the battery emf, battery voltage multiplied by 1 minus e -raise 2 minus t upon tau right here v is the battery v is the potential of the capacitor at any moment of time while e is the battery voltage or the maximum potential which the capacitor can have right here tau is the tau is the time constant of charging of capacitor through resistance or we can call it a time constant of a rc circuit which is nothing but value of resistance multiplied by value of capacitance now here the battery voltage is 10 volt and they are saying after a time interval of 2 milliseconds the capacitor voltage is raises to 0 .7 volt.

01:39 This is the data given.

01:43 So we put all the values v equal to, sorry, it is 7 volt, not 0 .7, 7 volt.

01:53 So we write here 7 is equal to 10, 1 minus e raise 2 minus t by tau, right? 7 by 10.

02:11 7 by 10 is 0 .7.

02:13 So it means we can solve it further here.

02:19 1 minus e raised 2 minus t by tau is equal to 7 by 10 means 0 .7.

02:29 We further solve it.

02:31 E raised to minus t by tau is equal to 1 minus 0 .7, is 0 .3.

02:42 Right? now here we take we have the equation here we have the equation here e raised to t by tau is equal to 0 .3 right it means we take we take log we take natural log right we take natural log on both side we take natural log on both sides we take natural log on both sides.

03:23 Right so l n e e raised to minus t by tau is equal to ln 0 .3 now we know the formula that log to the base e is to x is equal to x right we apply this formula here so this will become log e raise to log base e e e raise to x means x it means it is minus t by tau is equal to natural log of 0 .3 it means it is minus 1 .20 this gives us the value of tau as natural log of 0 .3 is minus 1 .20.

04:30 So this minus and this minus will get cancelled out.

04:35 So tau equal to t upon 1 .220 and we know the t is 2 milliseconds.

04:43 So it is 1 .20.

04:47 So it means it is 1 .66 into 10 raised to minus 2.

04:55 3 second is the answer for part a.

05:05 This is the answer for part a...

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