Solving 1st Order Circuits: Private Solution & Finding C, D

In summary, the conversation discusses a problem with a private solution for a first order circuit. The private solution is i_p = Bcos(wt + phase) and when placed in the equation, it becomes f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt+phase). The question is how to determine the values of C and D in this equation. The solution involves developing Acos(\omega t + \phi) into A_1cos(\omega t) + A_2sin(\omega t), where f[C,D] and g[C,D] represent the coefficients A_1 and A_2.
  • #1
asi123
258
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Ok, let's say I have a first order circuit with i_s = Acos(wt+phase) (like in the pic).
I'm having problem with the private solution, let's say I pick my private solution like this:
i_p = Bcos(wt + phase) = Ccos(wt) + Dsin(wt) (right?)
After I place this i_p in the equation I come up with this:
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt+phase) (right?)
Can I say
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt)?
I mean, How do I come up with C and D?

10x in advance.
 

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  • #2
I'm afraid I'm not understanding your question. What is i_p, f[C,D} and g[C,D]?
 
  • #3
asi123 said:
Ok, let's say I have a first order circuit with i_s = Acos(wt+phase) (like in the pic).
I'm having problem with the private solution, let's say I pick my private solution like this:
i_p = Bcos(wt + phase) = Ccos(wt) + Dsin(wt) (right?)
After I place this i_p in the equation I come up with this:
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt+phase) (right?)
Can I say
f[C,D]*cos(wt) + g[C,D]*sin(wt) = Acos(wt)?
I mean, How do I come up with C and D?

10x in advance.

You must develop [tex]Acos(\omega t + \phi) = A_1cos(\omega t) + A_2sin(\omega t)[/tex]

Now, [tex]f[C,D] = A_1[/tex] and [tex]g[C.D] = A_2[/tex]
 

1. What is a 1st order circuit?

A 1st order circuit is an electrical circuit that contains only one energy storage element, such as a capacitor or an inductor. These circuits can be analyzed using first-order differential equations.

2. How do I find the private solution for a 1st order circuit?

To find the private solution for a 1st order circuit, you will need to use the initial conditions of the circuit, such as the initial voltage or current across the energy storage element. These initial conditions will help you solve for the private solution using differential equations.

3. What is the significance of finding C and D in a 1st order circuit?

Finding the values of C and D in a 1st order circuit is important because they represent the constants in the general solution of the differential equation. These constants can tell you about the behavior of the circuit over time and help you find the complete solution.

4. Can I use the same methods to solve all 1st order circuits?

No, the methods used to solve 1st order circuits may vary depending on the type of circuit and the specific components involved. However, the general approach of using differential equations and initial conditions remains the same.

5. What are the applications of solving 1st order circuits?

Solving 1st order circuits is important in understanding the behavior of electrical circuits and can be applied in various fields such as electronics, telecommunications, and power systems. It is also a fundamental concept in circuit analysis and is essential for designing and troubleshooting circuits.

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