Respuesta :
Answer:
[tex]q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}[/tex]
at t = 0.001 we have
[tex]q = 1.55 \times 10^{-6} C[/tex]
at t = 0.01
[tex]q = 2.99 \times 10^{-6} C[/tex]
at t = infinity
[tex]q = 3 \times 10^{-6} C[/tex]
Explanation:
As we know that they are in series so the voltage across all three will be sum of all individual voltages
so it is given as
[tex]V_r + V_L + V_c = V_{net}[/tex]
now we will have
[tex]iR + L\frac{di}{dt} + \frac{q}{C} = 12 V[/tex]
now we have
[tex]1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12[/tex]
So we will have
[tex]q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}[/tex]
at t = 0 we have
q = 0
[tex]0 = 3\times 10^{-6} + c_1 + c_2 [/tex]
also we know that
at t = 0 i = 0
[tex]0 = -4000 c_1 - 1000c_2[/tex]
[tex]c_2 = -4c_1[/tex]
[tex]c_1 = 1 \times 10^{-6}[/tex]
[tex]c_2 = -4 \times 10^{-6}[/tex]
so we have
[tex]q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}[/tex]
at t = 0.001 we have
[tex]q = 1.55 \times 10^{-6} C[/tex]
at t = 0.01
[tex]q = 2.99 \times 10^{-6} C[/tex]
at t = infinity
[tex]q = 3 \times 10^{-6} C[/tex]
