An R L C series circuit has a voltage source given by E(t)=40 cos2 t V, a resistor of 2 Ω, an in

step 1 This question were to find the charge on the capacitor. Now, we're given a voltage source, E of

An R L C series circuit has a voltage source given by...

Key Concepts

Forced Oscillations and Sinusoidal Forcing

Many RLC circuits are analyzed under sinusoidal forcing conditions, where the source voltage is a cosine or sine function. This forcing leads to steady-state oscillations at the same frequency as the input source. In the solution process, a particular solution is found that directly addresses the sinusoidal input, which is then combined with the homogeneous solution that captures the circuit’s natural dynamics.

RLC Series Circuit Analysis

This concept involves understanding how resistors, inductors, and capacitors interact when connected in series in an electrical circuit. The analysis typically employs Kirchhoff's voltage law to derive a differential equation that models the circuit’s behavior. The system's response is a combination of transient effects due to stored energy in the inductor and capacitor and the influence of an applied voltage source.

Second Order Differential Equations

In the context of RLC circuits, the circuit’s behavior is described by a second order linear differential equation. This equation results from combining the contributions of the resistor (first order term), inductor (derivative term), and capacitor (integral or algebraic term). Solving such equations requires techniques that separately address the homogeneous (natural) response and the particular (forced) response, ensuring that the overall solution satisfies both the differential equation and the given initial conditions.

Application of Initial Conditions

When solving the differential equation for an RLC circuit, initial conditions such as the initial current and the initial charge on the capacitor are crucial. These conditions allow for the determination of the arbitrary constants in the homogeneous solution, ensuring that the total solution accurately reflects the physical state of the system at the beginning of the analysis.

Transcript

00:01 This question were to find the charge on the capacitor.

00:03 Now, we're given a voltage source, e of t, equal to 40 cosine of 2t.

00:11 Then using the notation in chapter 3 .5, we have an inductor l of 1 over 4 areas.

00:20 And then we have a resistor r, which is equal to 2 oms.

00:27 And then we have lastly a capacitor c, which is 1 over 13 ferrets.

00:37 Now, recall that from your testbook you have that e of c equal to q over c, er equal to ri, and then el equal to l, d -i -t.

00:54 And then when you use catch -off's law, that is using catch -off's law, when we use catch -off's law, we have that el plus er plus ec equal to e.

01:12 So when we write that down in mathematical form, we have that l, d -i, d -t plus r -i, plus i'll write this as 1 over c -q equal to e of t.

01:30 Now let me name this equation equation 1.

01:34 Going to come back to it again.

01:36 Now, recall that we have that recall that i equal to the q, the t.

01:46 And then the goal is to find what will be key of t, which is the charge and the capacitor.

01:51 Now, this would imply that the i, the t, when i differentiate both sides, will be equal to d squared of q over the t squared.

02:00 Right.

02:01 So that means that i can rewrite equation 1 as, so we can rewrite, we can rewrite equation 1 as, let's see what we can write that as.

02:16 So that will be l, my di d2 is not d squared q over dt squared plus r times i.

02:29 My i is the q d t and then plus one over c q equal to e of t now this is the equation we're going to be working weight and let me name this as equation two now let's continue solving this now that if i plug in the values i would have uh my l is one over four so i have one over four of d squared k over d t squared plus my r is two that's two d q d t plus one over c that would be one over one over 13 that's 13 q equal to 40 cosine of two c now let's multiply both sides by four we have that d squared q over the t squared plus 8 the q the t plus 52 q plus 52 q equal to 16 cosine of 2t now the homogeneous equation because we are going to solve this we're going to first of us solve the homogeneous parts so the homogeneous equation of let me name this equation equation equation we're going to come back to it later so the homogeneous equation of 3, it's as the auxiliary equation, as auxiliary equation, equation of, so this is the square.

04:04 So let's just use r.

04:05 That will be r squared plus its r plus 52 equal to 0.

04:12 Remember, we're solving the homogeneous part, so we equate that to 0.

04:16 Now, we solve this using the general formula, that will be r equal to minus b, which is.

04:22 Is minus 8 plus a minus square root of b squared, that's 64, minus 4 times my a is 1, my c is 52, divided by 2 times a, which is 2 times 1.

04:40 When you solve this, you have this is equal to minus 8 plus a minus square root of minus 144 divided by 2, which you can still write as minus 4 plus or minus of, yeah, square root of 144 is 12, 12 divided by 2, that would plus and minus 6i.

05:07 So that's r equal to minus 4 plus a minus 6i, because the i is coming because we have a minus here.

05:13 Now, this tells us that this implies that the q, almost.

05:20 Genius equation remember that we are solving for q will now be c1 of e to the because i have that minus four that will be minus 40 then cosine of 60 because i have 6 i and then plus c2 e to the minus 40 of sign of 60 now we need to find the particular solution so we use the method of undetermined coefficients, that's a method of undetermined coefficients, undetermined coefficients, i'll write coefficient as this.

06:00 So find the particular solution, to find the particular solution.

06:09 Now, what does the method of undetermined coefficient tell us? we'll guess that's because, recall that if we go back to this equation, we have from 3, we have 160 cosine of 2t.

06:20 So because we have cosine of 2 t so we can say that the q particular of t would be something like a cosine of 2t plus b sign of 2 t now we'll now plug this back we'll plug this back into equation 3 so um plug into equation 3 now before we can do that remember in equation 3 we need the squared q p d the t squared so let's of all first of all plug the let's find the q the t the q particular the t when you differentiate this you have minus two a sine of two t plus two b cosine of two t now when you look for d squared of q p d t squared you're going to have minus four a of cosine of two t minus four a of cosine of two t minus 4b of sine of 2t so now we plug this into 3 into equation 3 when you plug this into equation 3 you're going to have that um d squared q p d t squared up in minus 4a cosine of 2 t minus 4 b sine of 2 t and then you have plus a multiplied by the q the t that will be minus 2a sign of 2 t plus 2b cosine cosine of 2t and then we have a plus 52 multiplied by q that is a cosine of 2t plus b sign of 2t now this would equal 160 of 160 cosine of 2c now let's try and collect the like terms this would just when i simplify this you're going to have minus 4a plus 8 times 2 that's 16 so i'm collecting the cosine to that 16 b and then plus 52a multiplied by cosine of 2 t and they have plus let's collect the signs that will be minus 4b minus 16a comes from a and then plus 52b of sign 2t then this would be equal to 160 cosine of 2t now let's equates the cosines and the signs so this implies that i would have minus minus 4a plus 52a that is plus 58a so that implies that i have that i have that sorry that will be 48a so that means i have 48a plus 16b will be equal to because i have a cosine of 2c that will be equal to 160 and then for this minus 4b plus 52b so that will be minus 16a plus 48b equals to because i don't have a sign on the right hand side, this would be equal to zero.

09:53 Now from here, if i divide through by 16, i would have 3a plus b equal to 10.

10:04 And then this would imply that divide true, that implies that a equal to 3b.

10:11 Let me name this equation star and then this star star.

10:15 So what we do now is plug star into star star.

10:25 This implies that 3a, my a is now 3b plus b equal to 10.

10:34 That'll be 9b plus b.

10:36 That's 10b.

10:37 So this implies that b equal to 1.

10:41 And since b equals to 1, a equals to 3b.

10:44 So that implies that a will be equal to 3b...