A series circuit contains a resistor with R 24 ohms, an inductor with L 2 H, a capacitor with C 0.005 F, and a generator producing a voltage of E(t) 12 sin(10t). The initial charge is Q 0.001 C and the initial current is 0. Find the charge at time t.Q(t) ??
First thing to realize is that this circuit is in resonance. The characteristic frequency is equal to 1/sqrt(LC) 1/sqrt(2*.005) 10 This means the driving source won't provide net new charge to the capacitor and we can look at this in two pieces so for the output across the capacitor we have v-out 1/jωC/(R+jωL+1/(jωC))*v-in 1/(1-ω^2LC+jωRC) When we plug in values we get v-out1/(1-10^2*(2*0.005)+j10*24*.005) v-out1/(1-1+1.2j)*v-in -j/1.2*12sin(10t) 10sin(10t-π) so the change in Q across the capacitor from the driven source will be Q(t)-driven 0.005*10sin(10t-π) 0.0510sin(10t-π) C now we have to look at the initial conditions with a voltage of 0.2 V stored in the capacitor. We'll short out the driven source and solve for the decay in voltage through the resistor and inductor Because there is in essence a DC voltage stored on the capacitor and discharged through the resistor, the inductor has no role. The voltage discharge is 0.2exp(-t/RC)0.2exp(-8.33t). The charge is just this multiplied by C. or 0.2*0.005exp(-8.33t)0.001exp(-8.33t) the total will be Q(t)0.0510sin(10t-π)+0.001exp(-8.33t)
The differential equation for your circuit is: Lq'' + Rq' + q/C 12sin(10t).(1) If you assume that sin(10t) 0 for t0 you can find solution using one sided Laplace transform. I have assumed that sin(10t) is valid for -inf t inf . The general solution of non-homogeneous differential equation (1) consists of the sum of the particular (any) solution of the equation (1) and the general solution of the homogeneous equation (2) below: Lq'' + Rq' + q/C 0.(2) Particular solution of (1) is assumed as q Qsin(10t + fi). If you plug it in the solution is: q -0.05cos(10t).(3) The general solution of (2) is: q R*(e^at)*sin(bt + fi),.(4) where: a -R/2L -6 and b sqrt(4L/C-R^2)/2L 8. Plugging (4) into (2) you get R 0.051/0.8, fi 0.923. Finally your solution is the sum of (3) and (4): q - 0.05cos(10t) + (0.051/0.8)*(e^(-6t))*sin(8t + 0.923). You can verify that qo 0.001, and io q'(0) 0. Solution (2) is so called stationary solution. In this case LC is in resonance at frequency omega 10 and their serial impedance is 0. Therefore the stationary current is E/R 12/24 0.5 A, which you get as the first derivative of q. Solution (4) is the transient solution caused by initial charge q0.
