Question

Find the steady-state charge qp(t) and steady-state current ip(t) for the lrc-series circuit described by the given differential equation. find the complex impedance and impedance of the circuit. use the real method or complex method discussed on page 41 as instructed. d^2q/dt^2 + 2 dq/dt + 2q = 100 sin t.

a) 50 sin t, 2 sin t
b) 25 sin t, 4 sin t
c) 100 sin t, 2 sin t
d) 50 sin t, 4 sin t

Answer 1

To find the steady-state charge and current in an LRC-series circuit described by a given differential equation, we can solve the equation using the method of undetermined coefficients. The steady-state charge and current can be found by considering the homogeneous and particular solutions of the equation. The steady-state charge is -50sin(t) and the steady-state current is -50cos(t).

Step-by-step explanation:

To find the steady-state charge, we need to first solve the given differential equation. The equation is a second-order linear nonhomogeneous ODE with constant coefficients. We can solve it using the method of undetermined coefficients.

First, let's find the homogeneous solution by assuming a solution of the form qh(t) = Ae^(rt), where A is a constant and r is a variable. Substituting this into the differential equation, we get (r^2 + 2r + 2)Ae^(rt) = 0. This gives us the characteristic equation r^2 + 2r + 2 = 0. Solving this quadratic equation, we find that the roots are complex, which means the homogeneous solution is of the form qh(t) = e^(-t)(C1cos(t)+C2sin(t)).

Next, let's find the particular solution by assuming a solution of the form qp(t) = Dsin(t) + Esin(t), where D and E are constants. Substituting this into the differential equation, we get -2Dcos(t) + 2Esin(t) + Dsin(t) + Ecos(t) + 2(Dsin(t) + Ecos(t)) = 100 sin(t). Equating the coefficients of sin(t) and cos(t) on both sides, we can solve for D and E. We find that D = -50 and E = 0.

Therefore, the steady-state charge is qp(t) = -50sin(t) and the steady-state current is ip(t) = -50cos(t).