0:00
Hello.
00:02
So in this question it is given to us that a capacitor has an initial charge of q0 is discharged through a resistor.
00:11
What multiple of the time constant to gives the time the capacitor takes to loose a, the first 1 fourth of its charge and b, 3 fourth of its charge.
00:25
So in the first part of the question that is a, it is losing one -fourth of the total charge.
00:34
So it is losing one -fourth of q -0.
00:37
So the remaining charge, q is equal to q -0 minus one -fourth of q -0, which is equal to 3 -fourth of q -0.
00:52
Okay.
00:53
So this is the remaining charge.
00:56
Now the remaining charge in a capacitor whenever a capacitor discharges through a resistor is q is equal to q0 multiplied by e raised to minus t divided by toe where q0 is the original charge or the full charge, t is the time and toe is the time constant.
01:22
Okay, t is the time at which this charge has to be calculated.
01:26
This q has to be calculated.
01:28
Okay, so this q is equal to 3 fourth of q0.
01:31
So, 3 fourth of q0 is equal to q0, e raised to minus t divided by toe.
01:41
Okay, now this q0 and this q0 gets cancelled.
01:45
Now we have a e here.
01:47
Okay, there is a exponential value here.
01:49
So we will take log to the base of e on both sides.
01:55
Log to the base of e minus, lock to the base of e, e raised to minus t divided by toe is equal to log to the base of e 3 divided by 4.
02:11
Okay, that is how we will get rid of this exponential.
02:14
Okay, so minus t divided by toe is equal to this log to the log to the.
02:21
The base e and this e gets cancelled and this becomes minus t divided by two.
02:26
Okay, minus t divided by toe is equal to log to the base of e 3 divided by 4 is minus 0 .876...