The four capacitors in the circuit shown in the figure below, are connected across the terminals of a black box at t = 0.

The resulting current i_b for t > 0 is known to be: i_b = 50 e^-250t µA

If v_a(0) = 15 V,    v_c(0) = - 45 V,    v_d(0) = 40 V,   find the following for t ≥  0:

(a) v_b(t),  (b) v_a(t),  (c) v_c(t),  (d) v_d(t),  (e) i₁(t),  (f) i₂(t).

If C₁, C₂ and C₃ are three capacitances in series, then their equivalent C_eq will be obtained as follows:

          1/ C_eq = 1/ C₁ + 1/ C₂ + 1/ C₃ 

And If C₁ and C₂ are two capacitances in parallel, then their equivalent C_eq will be obtained as follows:

          C_eq  =  C₁ + C₂

Thus, equivalent capacitance in our circuit will be obtained as followed:

          C₁ = 1 + 1.5 = 2.5 nF

          1/C_eq =  1/ 2.5  +  1/12.5  +  1/50  =  ½   è  C_eq = 2 nF

Thus, we have: v_b(0)  =  v_a(0) + v_d(0) – v_c(0)  =  15 + 40 + 45 = 100 V

a)    V_b(t) =  - 1/C_eq  ∫^t₀ i_b dx  + v_b(0)  =  - 10⁹/2   ∫^t₀ 50. 10^-6 e^-250x dx  + 100
         =  - 25 000 e^-250x/(-250) |₀^t + 100
         =  100 (e^-250t - 1) + 100
         =  100 e^-250t  V

b)    V_a(t) =  - 1/C_a  ∫^t₀ i_b dx  + v_a(0)  =  - 10⁹/12.5   ∫^t₀ 50. 10^-6 e^-250x dx  + 15
         =  - 4 000 e^-250x/(-250) |₀^t + 15
         =  16 (e^-250t - 1) + 15
         =  16 e^-250t – 1  V

c)    V_c(t) =  1/C_c  ∫^t₀ i_b dx  + v_c(0)  =   10⁹/50   ∫^t₀ 50. 10^-6 e^-250x dx  - 45
         =   1 000 e^-250x/(-250) |₀^t - 45
         =  - 4 (e^-250t - 1) - 45  
         =  - 4 e^-250t – 41  V

d)    V_d(t) =  - 1/C_d  ∫^t₀ i_b dx  + v_d(0)  =  - 10⁹/2.5   ∫^t₀ 50. 10^-6 e^-250x dx  + 40
         =  - 20 000 e^-250x/(-250) |₀^t + 40
         =  80 (e^-250t - 1) + 40
         =  80 e^-250t – 40  V

CHECK:    v_b   =  v_a + v_d – v_c
                         =  16 e^-250t – 1  +  80 e^-250t – 40  - (- 4 e^-250t – 41)
                         =  100 e^-250t  V  (checks)

     e) i₁ =  - C_{(1 nF)} d(v_d)/dt
             =  - 10^-9 d(80 e^-250t – 40)/dt
             =  - 10^-9 (- 20 000 e^-250t)
             =  20 e^-250t µA

     f) i₂ =  - C_{(1.5 nF)} d(v_d)/dt
             =  - 1.5  10^-9 d(80 e^-250t – 40)/dt
             =  - 1.5  10^-9 (- 20 000 e^-250t)
             =  30 e^-250t µA

          CHECK:  i₁ + i₂ =  20 e^-250t  +   30 e^-250t  =  50 e^-250t µA  =  i_b