Ldi/dt + ir -Vcos wt=0 what technique do i use to solve?
RL AC circuit
- JSAMLv 51 decade agoFavorite Answer
You need to use an integrating factor to solve this differential equation. First re-write the equation as:
di/dt + i(R/L) = (V/L)*cos(wt)
On the left hand side, you see that we have the original variable 'i' and the derivative. To solve for i(t), we need to re-write the LHS into a derivative form (di/dt) and integrate. We need to define an integrating factor to do this. Let's define the integrating factor as:
exp(int[R/L dt]) where 'int' denotes integral. Also, let's denote it as 'u' to make simplification easier.
Next, multiply the above equation by 'u':
u(di/dt) + i*u(R/L) = u*(V/L)*cos(wt)
The 2nd term of the left hand side can be re-written as i*(du/dt). You can check this by checking the derivative manually:
u = exp(int[R/L dt])
du = exp(int[R/L dt])*(R/L)--->need to use Chain rule--->= u*(R/L)
Thus, we have:
u(di/dt) + i(du/dt) = u*(V/L)*cos(wt)
By inspection, you see that the LHS represents the product rule of the two functions 'u' and 'i'.
d(i*u)/dt = u*(V/L)*cos(wt)
Now you can integrate and solve for i(t):
i(t) = int[u*(V/L)*cos(wt)]/u where u = exp(int[R/L dt])
You could try to solve for the integral manually, but that would be rather difficult
Hope this helps
- yasiru89Lv 61 decade ago
di/dt + (r/L)i = (V/L)cos wt
When in this form we can find the integrating factor u(x) where
u(x) = exp [integral (r/L) dt]
Now note how by multiplying throughout by u(x) the LHS can be made exact and an application of integration on the RHS gives the general solution.
Hope this helps!