00:01
Hello, in the question we have given that a 3 -volt battery is connected to, so we have given the voltage as equal to 3 volts and it is connected to a capacitor whose capacitance is 8 .7 micro, sorry, millifarads and a 250 -ohm resistance.
00:22
So the resistance is given that is equal to 250 ohms.
00:28
So this is the value of resistor.
00:30
So all these are connected in series.
00:31
So how long does it take the battery to charge 95 % of capacity? so we have to tell, so in the part a, we have to find out the time.
00:43
So here we will be doing in this way.
00:46
So the first one is we have to find out the time taken.
00:50
So we will use the charging of capacitor over here.
00:54
So charging of capacitor.
00:57
So we have the formula that is q is equal to q naught times 1 minus e raised to minus t by tau, where tau is the time constant.
01:12
So let us first find out this tau.
01:14
Tau is nothing but rc that is equals to 250 into 8 .7 into 10 raised to minus 3.
01:23
So if we calculate this tau, so it turns out to be, so it will be 2 .175 seconds.
01:34
So now if we plug over here, so q is 0 .95 times q naught that is equals to q naught.
01:45
So 95%, so it will be 95 divided by 100 of q naught.
01:50
So this will be 0 .95 only.
01:52
So 1 minus e raised to minus t by tau is 2 .175 seconds.
02:00
So this will, this q naught gets cancelled.
02:03
And if i send this one over here, so i get this e raised to minus t by 2 .175 and that should be equal to, so 0 .05.
02:18
Now if i take ln on both the sides, that is natural log, so i will get minus t divided by 2 .175 that is equals to ln of 0 .05.
02:31
So if i calculate, i get minus t divided by 2 .175 that is equals to, so this is minus 2 .995.
02:42
So this minus gets cancelled and we get this t as equals to, so this t we will get it as 2 .995 into 2 .175.
02:56
So if we calculate this, the time, we will get it as 6 .51 seconds.
03:03
So this is the time taken to charge the capacitor...