00:01
In the question, it is given that, given that capacity of tank 1 which is c1 is equal to 10 multiplied by a and capacity of tank 2 that is c2 is equal to 10 multiplied by b and resistance of tank 1 that is r1 is equal to 1 and resistance of tank 2 that is r2 is equal to 4.
00:54
So here a is equal to 10 and b is equal to 7.
00:58
We can use volume properties and the definition of resistance to write the following equations for tank 1.
01:04
So for tank 1 we have c1 dh1 upon dt is equal to 2 minus q1 where q1 is equal to h1 upon r1.
01:19
So comparing this gave us the first equation of motion that c1 dh1 upon dt is equal to q minus h1 upon r1.
01:32
Let us take this as equation 1.
01:33
If we will do the same for the tank 2 then we will get the following equation like c2 dh2 upon dt is equal to q1 minus q2.
01:46
So c2 dh2 by dt will be equal to h1 upon r1 minus q2.
01:56
Let us take this as equation 2.
01:59
Now taking the transform for both the equation of motion keeping in mind that all initial conditions are 0.
02:14
So equation 1 will become c1sh1 of s is equal to q of s which is q of s minus 1 upon r1 h1 of s.
02:31
Now equation 2 will become c2sh2 of s is equal to 1 upon r1 h1 of s minus q2 of s.
02:48
So we need to eliminate h1 of s from these equations.
02:53
So we can initially solve the first equation for h1 of s.
02:58
So c1sh1 of s is equal to q of s minus 1 upon r1 h1 of s.
03:13
So c1sh1 of s plus 1 upon r1 h1 of s will be equal to q of s.
03:23
So on taking h1 of s as common we will have c1 of s plus 1 upon r1 in the bracket which is equal to q of s...