Question #107018
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
2020-04-02T13:05:52-0400

The magnitude of the subsequent current can be obtained from the equation of damped oscillations : q¨+2βq˙+ω02q=0\ddot{q}+2\beta\dot{q}+\omega_0^2q=0 , where β=R2L,ω0=1LC\beta=\frac{R}{2L},\omega_0=\frac{1}{\sqrt{LC}}

In our case ω02β2<0\omega_0^2-\beta^2<0 then is a case of aperiodic damping. The equation solution is

q=c1eα1t+c2eα2tq=c_1e^{-\alpha_1t}+c_2e^{-\alpha_2t} where α1=β+β2ω02\alpha_1=\beta+\sqrt{\beta^2-\omega_0^2} , α2=ββ2ω02\alpha_2=\beta-\sqrt{\beta^2-\omega_0^2}

From the conditions q(0)=q0=c1+c2q(0)=q_0=c_1+c_2 and q˙(0)=I(0)=0=α1c1+α2c2\dot{q}(0)=I(0)=0=\alpha_1c_1+\alpha_2c_2 we find

c1=α2q0α1α2c_1=-\frac{\alpha_2q_0}{\alpha_1-\alpha_2} ,c2=α1q0α1α2c_2=\frac{\alpha_1q_0}{\alpha_1-\alpha_2} then q˙(t)=I(t)=q0α1α2α1α2(eα1teα2t)\dot{q}(t)=I(t)=q_0\frac{\alpha_1\alpha_2}{\alpha_1-\alpha_2}(e^{-\alpha_1t}-e^{-\alpha_2t}) . As α1=50,α2=10\alpha_1=50, \alpha_2=10 then

I(t)=12.5q0(e50te10t)I(t)=12.5q_0(e^{-50t}-e^{-10t}) A


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