Answer to Question #107204 in Differential Equations for Gayatri Yadav

Question #107204
A series RLC circuit with R = 6 ohm, C = 0.02 Farad and L = 0.1 has no applied voltage. Find the subsequent current in the circuit if the initial charge, on the capacitor is q0 and the initial current is zero.
1
Expert's answer
2020-04-01T10:20:30-0400

A series "RLC" circuit with "R=6" ohm, "C=0.02" Farad and "L=0.1" has no applied voltage. Find the subsequent current in the circuit if the initial charge, on the capacitor is "q_0" and the initial current is zero. 

Kirchhoff's voltage law:

"u_R+u_L+u_C=0"

where "u_R,u_L,u_C" are the voltages across "R,L" and "C" respectively.

Substituting in the constitutive equations:

"Ri(t)+L\\frac{di(t)}{dt}+\\frac{1}{C}\\int \\limits_{-\\infty}^{t}i(\\tau)d\\tau=0"

Differentiating and dividing by "L" :

"\\frac{d^2i(t)}{dt^2}+\\frac{R}{L}\\frac{di(t)}{dt}+\\frac{1}{LC}i(t)=0"

This can usefully be expressed in a more generally applicable form:

"\\frac{d^2i(t)}{dt^2}+2\\alpha\\frac{di(t)}{dt}+\\omega_0^2i(t)=0\\\\\n\\alpha=\\frac{R}{2L}, \\omega_0=\\frac{1}{\\sqrt{LC}}"

The differential equation has the characteristic equation: 

"s^2+2\\alpha s+\\omega_0^2=0"

The roots of the equation in "s" are:

"s_{1,2}=-\\alpha\\pm\\sqrt{\\alpha^2-\\omega_0^2}"

The general solution of the differential equation is an exponential in either root or a linear superposition of both

"i(t)=Ae^{s_1t}+Be^{s_2t}"

The initial current is zero: 

"i(0)=Ae^{0}+Be^{0}=0\\\\\nA=-B"

"i(t)=A(e^{s_1t}-e^{s_2t})=\\\\\n=Ae^{-\\alpha t}(e^{\\sqrt{\\alpha^2-\\omega_0^2}t}-e^{-\\sqrt{\\alpha^2-\\omega_0^2}t})\\\\\n\\alpha=\\frac{6}{0.2}=30, \\omega_0=\\frac{1}{\\sqrt{0.002}}\\\\\n\\sqrt{\\alpha^2-\\omega_0^2}=\\sqrt{900-\\frac{1}{0.002}}=\\sqrt{400}=20"

therefore

"i(t)=Ae^{-30 t}(e^{20t}-e^{-20t})=A(e^{-10t}-e^{-50t})"

The initial charge on the capacitor is "q_0" and initial current is zero: 

"L\\frac{di(t)}{dt}|_{t=0}+\\frac{q_0}{C}=0\\\\\n\\frac{di(t)}{dt}|_{t=0}=A(-10e^{-10t}+50e^{-50t})\\\\\n\\frac{di(t)}{dt}|_{t=0}=40A\\\\\n\\frac{q_0}{LC}=-40A\\\\\nA=-\\frac{q_0}{40LC}=-12.5q_0"

Therefore:

"i(t)=-12.5q_0(e^{-10t}-e^{-50t})"



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