A 12.0 ?F capacitor is charged to a potential of 50.0 V and...

A 12.0 ?F capacitor is charged to a potential of 50.0 V and then discharged through a 225 ? resistor. How long does it take the capacitor to lose half it's charge? 3.4 x 10^-3 s 4.3 x 10^-3 s 1.87 x 10^-3 s .01 s

Transcript

00:01 This question we are given a capacitor so we are given a capacitor c which is equal to capacitance here is 12 micro ferrari which is equal we can write 12 multiple multiple minus 6 ferrari and the charge so we are given a potential v which is equal we can write 50 volt potential is 50 volt and the discharge tube have a resistance so resistance we are given here 225 ohm and how long does it take to capacitor lose the half of its charge? so we want to find value of time t so let's solve this question and finding the value of time t here so first we can write the value of capacitor the discharge so here we can write v which is equal to v0 multiple we can write a 2 minus t divided by tau this is our formula we are we are given here and now we know that tau which is equal to our multiple over c which is equal we can write 200 and here we can write 225 they are given here 225 multiple by we can write value of c is 12 multiple by 10 dash to minus 6 12 multiplied by n x which is equal to answer here is so here we can write after simplification we can write 225 multiple by 12 which is equal to answer is that 2700 by 10 x to minus 6 second substitute value into this formula so we can write formula v which is equal v0 multiple we can weigh 2 t divided by tau now we just substitute value into this formula.

01:34 Value into this formula so here we can write the value is 50 so which means original we are given here that voltage 50 that is half of this voltage and here we can wait which means is v is v0 divided by 2 which is 25 and here we can write e ratio t divided by 2700 multiplied now we just simplify and we can write here answer is we can write 1 divided by 2 which is equal we can write e -dh -2 1 divided by we can write here 2 .7 here we can write 2 .7 milliseconds now again we can write here taking here we are simply find and we can write that minus t divided by 2 .7 which is equal we can write l .n of 0 .5 now we just equal to we can write here answer is 0 .693 here we can write plus we just put in minus on the right -hand side now again we just simplify and we can write to t which is equal to 2 .7 multiple by we can write here 0 .63 and putting here minus and we know the time is positive so we can write plus now here we have positive this value is positive is negative we know that ln 0 .5 which is equal to minus 0 .693 now here after simplification we can write here answer is so here 2 .7 multiplied by 0 .693 which is equal to answer is 1 .87.

03:10 So answer here is 1 .87 milliseconds...

Question

A 12.0 ?F capacitor is charged to a potential of 50.0 V and then discharged through a 225 ? resistor. How long does it take the capacitor to lose half it's charge? 3.4 x 10^-3 s 4.3 x 10^-3 s 1.87 x 10^-3 s .01 s

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