Question

A circuit consists of a 6 ohm resistor, a 0.3 farad capacitor, and an AC voltage source supplying V(t) = 120 cos(10 t) volts. 1. Write the differential equation for the charge on the capaciton q'+0.55 0.56 q=20 cos(10t) X 2. Write a formula for the solution, assuming q(0) 0 q(f) =

Answer 1

Apply the Kirchhoff's loop law to the RC circuit to obtain the loop equation. The expression for the Kirchhoff's loop equation for the RC circuit is, -iR-9199 +V(t) = 0 с The expression for the current through the circuit is, I i= dg (t) dt - The expression for the time dependent voltage is, V(t) = 120 cos(10) The expression for the linear differential equation of first order and its integrating factor is, dy +Py=Q Integrating factor (1F) = el Par solution = y(LF) = S Q(IF)+C

Then write the linear differential equation of first order for charge as follows: -iR-9(t) +V(t) = 0 (dal) R+90) =v (1) dg() 16 90)= dt *(69)03719(t) = 120 cos(107) 622 dq(t) dt *(1.85)9(t)= 20cos(10t) (1.85): Q = 20cos(10t) The integrating factor for the above differential equation is, Fue zelés a

The solution for the above differential equation is, q(t)(IF)= (20 cos (101))(IF)dt+C 9(0)= c., (5 (20cos (10)([F]dt +C) (0) = 20 (sair cos (10) d +C) .... (1) The expression for the integration formula is, s ea cos(bt)dt = 2 (acosbt +b sin bt) ...... (2) Then compare the equations (1) and (2) as follows: Ot) +10 sin (10t) +C 1.85 e to +6037/(1.85 )cos(101) +10 sin (101)+ COS(10t) + 10 sin (10) + C BS) =(021 / . cos(101): 10sin(10) 200 01.8

Assume q(0) = 0 to fine out the constant value C. 90-(021( 1. cos (10)-tosin (10) 2012 9(0)=(0.2) ( 155 )cos(1000) +10sin (1000)) 200 0=(02) (15. )(a)+0)200 C--180 Therefore, the expression for the charge is, 9(6)=(02) ( 135 )cos(105) + 10 sin(101): 200 960)=(02) . ) cos(10)+10in (10) - 150 90=(1cos(106) + 2 sin(101) )