Acircuit consists of a 6 ohm resistor, a 0.2 farad capacitor, and an ac voltage source supplying v(t) = 120 cos(20 t) volts. write the differential equation for the charge on the capacitor. write a formula for the solution, assuming q(0) = 0.

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))

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the ima (higher than 1m) will be greater than the ima with the height of 1m.