The question :

Find Suppos C = 2.5 x 10-4, Q(0) = 0, and E(t) = 100 cos 120t. (a) Find Q(t) and I(t). (b) What is the amplitude of the the steady-state current? on 1 10. An emf of voltage E(t) = Eo coswt is applied to the RC Find circuit of Fig. 3.7.8 at time t = 0 (with the switch closed), and O(0) = 0. Substitute esp(t) = A cost + B sin wt in with the differential equation to show that the steady periodic of charge on the capacitor is me, esp (1 ) = EOC V1 + 0 2 R 2 C2 cos(wt - B) on nd where B = tan (WRC). for In Problems 11 through 16, the parameters of an RLC circuit 1. with input voltage E(t) are given. Substitute nd Isp(t) = Acos wt + B sin wt in Eq. (4), using the appropriate value of w, to find the steady periodic current in the form Isp(t) = lo sin(wt - 8). 11. R = 30 02, L = 10 H, C = 0.02 F; E(t) = 50 sin 2t V 12. R = 200 32, L = 5 H, C = 0.001 F; E(t) = 100 sin 10t V 13. R.= 20 02, L = 10 H, C = 0.01 F; E(t) = 200 cos 5t V 14. R = 50 0, L = 5 H, C = 0.005 F; O E(t) = 300 cos 100t + 400 sin 100t V 15. R = 100 92, L = 2 H, C = 5 x 10-6 F; E(t) = 110 sin 60rt V 16. R = 25 02, L = 0.2 H, C = 5 x 10-4 F; E(t) = 120 cos 377t V3 Linear Equations of Higher Order If we substitute (1) in Eq. (2), we get the second-order linear differential equation LQ" + Re't- e= E(t ) (3) for the charge Q(t), under the assumption that the voltage E(t) is known. In most practical problems it is the current / rather than the charge Q that is of primary interest, so we differentiate both sides of Eq. (3) and substitute I for Q' to obtain LI" + RI'+ - I = E'(t). (4) We do not assume here a prior familiarity with electrical circuits. It suffices to regard the resistor, inductor, and capacitor in an electrical circuit as "black boxes" that are calibrated by the constants R, L, and C. A battery or generator is described by the voltage E(t) that it supplies. When the switch is open, no current flows in the circuit; when the switch is closed, there is a current I (t) in the circuit and a charge Q(t) on the capacitor. All we need to know about these constants and functions is that they satisfy Eqs. (1) through (4), our mathematical model for the RLC circuit. We can then learn a good deal about electricity by studying this mathematical model. The MechanicIn Problems through, the parameters of an RLC circuit with input voltage E(t) are given. Substitute Isp(t) = A cos wt + B sin wt in Eq., using the appropriate value of w, to find the steady periodic current in the form /sp(t) = /0 sin(wt - 6 ). R = 50 0, L = 5 H, C = 0.005 F; E(t) = 300 cos 100t + 400 sin 100t VIn the problem, the parameters of an RLC circuit with input voltage E(t) are given. Substitute Is p (t) = Acosoot + Bsindot in the equation : LI" + RI' + (1/C)I = E'(t) using appropriate values of w, to find the steady periodic current in the form I_sp (t) = 1_0 sin(wt - 6) (i think o is delta?) R = 100 Ohm L = 2H C = 5 * 10-F E(t) = 110sin(60 m t) V the answer should be: Is p (t) = 0.1591 sin(60TI t - 4.8576) approximately