00:01
Hello, in the question we have to find the charge on capacitor in the rlc series circuit.
00:07
So this is the circuit and these are the value of resistor, then inductor and the capacitor.
00:13
So now in this circuit we will just apply kvl.
00:17
So applying kvl, so see this current flows in this direction.
00:22
So first i encounter minus, so it is minus of at, then minus of ir plus c.
00:29
This is emf.
00:30
So potential drop across this is ldi by dt across the inductor and for this capacitance, so q is equal to cv, so v will be equal to q by c.
00:41
This is the potential drop across this capacitance and this is equal to zero.
00:47
But now i know that the current is given by dq by dt.
00:52
So wherever i see i, i will just put dq by dt.
00:56
So doing so, so here it is i, so it will be d2q by dt square plus r dq by dt plus q by c is equal to e of t.
01:07
Now see, i will have just put the value of l, which is 1 by 5, r is 2 by 5 and c is 1 by 2.
01:15
So it is 2q that is equal to e of t is 50.
01:19
Now this is a differential equation.
01:22
So we will solve it by using the condition.
01:28
So we have cf.
01:31
So complementary function it is d2q plus, so what i have done is i have taken, so this 5 i have multiplied over and i got this.
01:44
So i made it first homogeneous.
01:47
So i am just calculating cf.
01:50
So cf will be d2q by dt square plus 2q by dt plus 10q.
01:56
So this 5 i have multiplied on the both the sides.
02:02
So, so this is cf.
02:04
Now auxiliary equation will be d2q plus 2dq plus 10q.
02:12
So i am taking q common.
02:14
So it will be this, this is equal to zero where what is d? d is the differential operator that is d by dt and d2 is d2 by dt square.
02:25
So now this is my equation.
02:31
So this is a quadratic type equation.
02:33
So if we solve it, so minus b plus or minus square root of b square minus 4ac.
02:40
So by using that formula, so we get this as, so solving this quadratic equation.
02:49
So i have solved this.
02:51
So b square is 2.
02:53
So this number over here turns out to be imaginary because it is minus of 36.
02:59
So square root of minus of 36 is 6 iota.
03:02
So i got this.
03:04
So d will be equal to minus 2 plus or minus 6 i divided by 2.
03:09
So if i further solve this, i will get this d as equal to plus or minus 1 plus or minus 3 i.
03:17
Now this is an complex root.
03:21
So this is a complex root and for the complex root, the complementary function, the solution for the equation is given by e raised to the imaginary part, sorry the real part times t into a cos of the imaginary part and b sin of the imaginary part.
03:40
So this is what it lands over here.
03:43
So q of t is equal to e raised to minus.
03:45
So real part is minus 1...