00:01
Hello students, let's solve this question here.
00:03
The voltage across the resistor r2 is v and is same as the voltage across a capacitor.
00:12
We need to get the capacitor voltage as v of t is equal to v of infinity plus v of zero plus minus v of infinity into e raised to minus t divided by r.
00:33
Now where tau is a time constant and v of infinity is a finite value and v of zero plus is a initial value.
00:40
So let's start by finding v zero plus by first analyzing at t is equal to zero minus.
00:53
So at t equal to zero, it is assumed to be at a steady state.
01:04
So at the steady state capacitor acts as zero coulomb.
01:12
So by voltage division we can write v zero minus will be equal to 17 k divided by 17 plus 19 into v s.
01:22
Therefore we'll get the v s value is 20 .2.
01:29
Therefore on simplification we'll get v zero minus will be equal to 17 into 20 .2 divided by 36 which will be equal to 9 .53 volts.
01:43
Now when the switch moves to the position b at t is equal to zero plus, the capacitor doesn't allow sudden change in voltage.
01:54
Therefore v of zero minus will be equal to v of zero plus.
02:00
Hence v of zero plus will be equal to 9 .53 volts.
02:08
Now let's find the time constant tau for that the equation tau is equal to the equivalent resistance into the capacitance where r eq is a thevenin equivalent resistance and it's seen by the capacitor.
02:25
So r2 and r3 are connected parallel in the circuit.
02:29
Therefore we can say that r equivalent is equal to r2 parallel r3 becomes r2 into r3 divided by r2 plus r3.
02:41
Let's substitute the values which is equal to 17 into 9 divided by 17 plus 9 and which will be equal to 5 .884 ohm.
02:51
And now it is given that the capacitance is equal to 178 nano farad...