The voltage and the current relationships in circuit analysis
"i= \\frac{dq}{dt}\n\u200b"
voltage drop across the inductor
"V_L =L\\frac{di}{dt}=L\\frac{d^2q}{dt^2}\u200b"
voltage drop across the resistor
"V_R=Ri=R\\frac{dq}{dt}"
voltage drop across the capacitor
"V _C=\\frac{q}{C}"
Kirchhoff's Second Law
"L\\frac{d^2q}{dt^2}+R\\frac{dq}{dt}+\\frac{q}{C}=E(t)"
Given that "L=1"H, "R=1000"ohm, "C=10^{-6}" F, "E(t)=12" V
"1\\frac{d^2q}{dt^2}+1000\\frac{dq}{dt}+\\frac{1}{10^{-6}}q=12\\\\\n\\frac{d^2q}{dt^2}+1000\\frac{dq}{dt}+1000000q=12\\\\"
The characteristic equation is
"r^2+1000r+1000000=0\\\\\nD=(1000)^2-4\\cdot1000000=-3000000\\\\\nr_1=\\frac{-1000+i1000\\sqrt3}{2}=-500+500\\sqrt3i\\\\\nr_2=\\frac{-1000-i1000\\sqrt3}{2}=-500-500\\sqrt3i\\\\\nq_C=e^{-500t}(c_1\\cos(500\\sqrt3t)+c_2\\sin(500\\sqrt3t))"
Find the particular solution of the non-homogeneous differential equation using the method of the undetermined coefficients.
Assume that "Q(t)=A" is a solution of the non-homogeneous differential equation where "E(t)=12"
Differentiate the assumption with respect to "t"
"Q'=0, Q''=0"
Substitute in the original equation
"0+0+1000000A=12\\\\\nA=1.2\\cdot10^{-5}\\\\\nQ(t)=1.2\\cdot10^{-5}\\\\\nq(t)= e^{-500t}(c_1\\cos(500\\sqrt3t)+c_2\\sin(500\\sqrt3t))+\\\\\n+1.2\\cdot10^{-5}"
The initial charge of the capacitor is zero
"q(0)= e^{-500\\cdot 0}(c_1\\cos(500\\sqrt3\\cdot 0)+\\\\\n+c_2\\sin(500\\sqrt3\\cdot 0))+1.2\\cdot10^{-5}=0\\\\\nc_1=-1.2\\cdot10^{-5}"
The circuit is closed at "t=0: i(0)=q'(0)=0\\\\"
"q'(t)= -500e^{-500t}(-1.2\\cdot10^{-5})(\\cos(500\\sqrt3t)-\\\\\n-500e^{-500t}(c_2\\sin(500\\sqrt3t))+\\\\\n+ 1.2\\cdot10^{-5}(-500\\sqrt3)e^{-500t}(\\sin(500\\sqrt3t)+\\\\\n+c_2\\cdot500\\sqrt3e^{-500t}\\cos(500\\sqrt3t))\\\\\nq'(0)=-500(-1.2\\cdot10^{-5})+c_2\\cdot500\\sqrt3=0\\\\\nc_2=-0.4\\sqrt3\\cdot10^{-5}\\\\\nq(t)=e^{-500t}(-1.2\\cdot10^{-5}\\cos(500\\sqrt3t)-\\\\\n-0.4\\sqrt3\\cdot10^{-5}\\sin(500\\sqrt3t))+1.2\\cdot10^{-5}"
Coulomb:
"q(1)=e^{-500\\cdot(1)}(-1.2\\cdot10^{-5}\\cos(500\\sqrt3\\cdot(1))-\\\\\n-0.4\\sqrt3\\cdot10^{-5}\\sin(500\\sqrt3\\cdot(1)))+1.2\\cdot10^{-5}\\approx\\\\\n\\approx1.2\\cdot10^{-5}(C)"
Find the steady state charge.
"t\\to\\infty:q_{p}(t)\\to1.2\\cdot10^{-5}C"
Comments
Leave a comment