Tuesday, April 20, 1999
Announcements:
The final exam will be 110 minutes long. It will consist of 25 multiple choice questions. 15 of the 25 will be on new material with the remaining 10 on material covered on the first and second midterm.
Lecture notes:
 Continued from last lecture...
 Solution for d²q/ dt² = -q /LC
    q = Q cos(wt + d) where d is the initial condition
    w = (1/ LC)^1/2

• Since we said the capacitor is charged to a maximum value of Q at t= 0, d = 0
• i = dq/dt = -Qw Sin (wt) = -i_max Sin (wt)
• U = U_E + U_B = q²/2C + (1/2)Li² = Q²/2C cos² (wt) + (1/2)L (i_max )² sin² (wt) =

• As current flows through the resistor, electrical energy is dissipated as heat
• Without R; U = U_E + U_B = q²/2C + (1/2)Li² = a constant
• With R:  dU/dt = -i²R = d/dt [Li²/2 + q²/2C]
• or...  -i²R = Li di/dt + q/c dq/dt
• -iR = L di/dt + q/c
• L di/dt + q/c + iR = 0
• L d²q/dt² + q/c + R dq/dt = 0
• Solution?
• q = Qe^-Rt/2L cos (w^lt + d)   Charge on a capacitor
• w^l = ( w² - (R/2L)²) ^1/2 where w = (1/ LC)^1/2
• U = Q²/2C e^-Rt/L

• We learned that by rotating a coil in a magnetic field, a sinusoidal emf can be generated. What if such a generator (instead of a battery) is placed in a circuit?

• Resistive load

• x = x_max sin wt = v_R
• Let's assume that instantaneously the current i_R is clockwise, i_R = v_R / R
• By the loop rule, x - v_R = 0, therefore v_R =V_R sin wt = x_max sin wt  and i_R = I_R sin wt  = (v_R / R)  sin wt

• v_R and i_R are said to be phase, T = one period
• Phasor Diagram

•  i_R = I_R sin wt  and v_R =V_R sin wt ( or the projection of the phasors on y axis)