Respuesta :
Answer:
q = 24 cos (20t) (1- e (-t / 1.2))
Explanation:
To work this circuit we use the mesh equation (Kirchoff) that states that the voltage in a closed circuit is zero
Vg + Vr + Vc = 0
Where Vg is the generator voltage, Vr the resistance voltage and Vc the capacitor voltage
Vg = 120 cos 20t = V
Vr = i R
Vc = q / C
We see that in one term we have the current (i) and in another the load (q), but there is a relationship between the two
i = dq / dt
Let's replace in the initial equation
V + R dq / dt + q / C = 0
Reorder the terms
Rdq / dt + q / C - V = 0
dq / dt + q / rC - V / R = 0
dq / dt = V / r -q / RC
dq / (V / R -q / RC) = dt
we integrate
∫I dq / (V / R - 1 / RC q) = ∫ dt
We change variables
u = (V / R - 1 / RC q)
du = -1 / RC dq
∫ dq / (V / R - 1 / RC q) = -RC ∫ du / u
-RC ln u = -RC ln (V / R - 1 / RC q)
We evaluate between the limits of integration, the lower t = 0 q (0) = 0
-RC [ln (V / R - 1 / RC q) - ln (V / R)] = t
[ln (V / R - 1 / RC q) - ln (V / R)] = -t / RC
Ln (V / R - 1 / RC q) / (V / R)] = -t / RC
Ln (1 - 1 / VC q)) = (-t / RC)
Ln (VC- q) = ln (VC) (-t / RC)
VC-q = VC e (-t / RC)
q = VC - Vc e (-t / RC)
q = VC (1- e (-t / RC)
We substitute the values they give us
q = (120 cos (20t) 0.2) (1- e (-t / 6 0.2))
q = 24 cos (20t) (1- e (-t / 1.2))
