A series circuit contains a resistor with R = 24 ohms, an inductor with L = 2 H, a capacitor with C = 0.005 F?

Question

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 12 V battery. The initial charge is Q = 0.001 C and the initial current is 0. Find the charge and current at time t.
Q(t)= ??

Answer 1

Ok, we need to solve the circuit, so then we need to use Kirchoff :

-IR - LdI/dt - q/c = 0

IR + LdI/dt + q/c = 0

dq/dt = I =

dq/dt*R + Ld^2q/dt^2 + q/c =0

d^2q/dt^2 + dq/dt*R /L + q / LC = 0

This is the differential equation we must solve :

( D^2 - R/L*D + 1/LC ) = 0

D = R/L +/- sqrt( (R/L)^2 - 4/LC) over 2

D = R/2L +/- sqrt( (R/2L)^2 - 1/LC)

where D es el valor de w', in other words the angular frequency

now : R/2L = X ; 1/LC = y

if y^2 - x^2 > 0 >>>>> D = R/2L + i*sqrt ( 1/LC - (R/2L)^2 )

D = R/2L - i*sqrt( 1/LC - (R/2L)^2 )

w' = sqrt(1/LC - (R/2L)^2) rad/s

q = Ae^(R/2L + i*(w'))t + Be^(R/2L - i*(w'))t

q = A*(e^R/2L*t)(e^i*(w't))

q = A*(e^R/2L*t)(cosw't + i senw't)

we know, using math :

q(t) = A*(e^R/2l*t)cos(w't+alfa)

replacing the values :

w' = sqrt( 1 / 2*0.005 - (24 / 4)^2)

w' = 8 rad/s

then :

q(t) = 0.001(e^6t)*cos(8t) C

Hope that helped

Answer 2

The differential equation to your circuit is: Lq'' + Rq' + q/C = 12sin(10t).............(a million) in case you assume that sin(10t) = 0 for t<0 you will discover answer using one sided Laplace rework. I unquestionably have assumed that sin(10t) is valid for -inf < t < inf . the final answer of non-homogeneous differential equation (a million) contains the sum of the actual (any) answer of the equation (a million) and the final answer of the homogeneous equation (2) under: Lq'' + Rq' + q/C = 0.............(2) particular answer of (a million) is theory as q = Qsin(10t + fi). in case you plug it in the answer is: q = -0.05cos(10t).............(3) the final answer of (2) is: q = R*(e^at)*sin(bt + fi),....(4) the place: 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 answer is the sum of (3) and (4): q = - 0.05cos(10t) + (0.051/0.8)*(e^(-6t))*sin(8t + 0.923). you may verify that qo = 0.001, and io = q'(0) = 0. answer (2) is so noted as table certain answer. for this reason LC is in resonance at frequency omega = 10 and their serial impedance is 0. subsequently the table certain cutting-edge is E/R = 12/24 = 0.5 A, which you get because of the fact the 1st spinoff of q. answer (4) is the quick answer led to by way of preliminary can charge q0.