8. [6 marks] Consider the model equation describing the flow...

8. [6 marks] Consider the model equation describing the flow of electric current in a simple RLC series circuit: $LI''(t) + RI'(t) + \frac{1}{C}I(t) = 0$. Given that the resistance $R$ is 0.25 ohms and the inductance $L$ is 1 henry, we need to determine the largest possible integer value of capacitance $C$ that will ensure the current $I(t)$ in the circuit continues to change direction (i.e., oscillate) as $t$ approaches infinity.

Transcript

00:01 In this question we are talking about rcl circuit which is connected in series and we are given the values for resistance it is 4 ohm for capacitance it is 1 over 26 faraday then inductance is equal to 1 by 2 henry then applied voltage e of t is equal to 16 sin 70 we have to find out the charge at any time t using the method of undetermined coefficient now since we are talking about the rcl circuit which is connected in series so the differential equation can be written in this form l d2q plus dt square plus r dq by dt plus 1 by c of q equal to e of t now putting the values this will be 1 by 2 d2q by dt square plus resistance is 4 dq by dt plus this will be equal to 26 q is equal to 16 sin 70 now multiplying both sides by 2 the differential equation will become like this this will become 52q and this will be 32 sin 70 now we have to solve this differential equation also we are given that there is no initial charge and no initial current no initial current means i of 0 is 0 and no initial charge means q of 0 is equal to 0 current is change in charge with respect to time so this can be written as q dash 0 equal to 0 as well these two will basically be our initial conditions now this is non -homogeneous differential equation so if we consider the corresponding homogeneous differential equation this will be d2q over dt square plus 8 dq by dt plus 52q is equal to 0 auxiliary equation will be m square plus 8m plus 52 equal 0 that means m will be equal to minus 8 plus minus square root of 8 square that is 64 minus 4ab this will be 4 into 52 which is 208 divided by 2a, a is 1 that means this will be minus 8 plus minus square root of minus 144 divided by 2 so m will be equal to minus 8 plus minus 12 iota over 2 or this can be written as minus 4 plus minus 6 iota so this is of the form alpha plus iota beta with alpha equal minus 4 and beta equal 6 that means the solution will be of the form e raised to power alpha t that means minus 4t into c1 cos beta t that means cos 60 plus c2 sin 60.

03:15 Now we will find out the particular solution qp since in the differential equation we have sin 17 terms so that means this will be a into cos 70 plus b into sin 70 qp dash will be equal to minus 7a sin 70 plus 7b cos 70 this is qp dash and similarly qp double dash will be minus 49a cos 70 minus 49b sin 70 putting these three values in the given differential equation so we will have minus 49a cos 70 minus 49b sin 70 plus 8 into minus 7a sin 70 plus 7b cos 70 plus 52 into qp.

04:32 That means a cos 70 plus b sin 70 equal to this is 32 sin 70.

04:46 Now on simplifying we will get that the coefficient of cos 70 is 56b plus 3a and the coefficient of sin 70 is 3b plus 56a equal to 32 sin 70.

05:07 On comparing we will have two equations first is 56b plus 3a is equal to 0 and second one will be 3b plus 56a is equal to 32.

05:22 From here we will get that b is a is equal to minus 56b by 3 putting this value here and on solving we will get the value of b is minus 96 over 3127.

05:39 Now putting this value here we will get that a is equal to 1792 divided by 3127...

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