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.

Expert's answer

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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