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FINAL EXAM PART 2 SP'12
QUIZ 10   ~   CH. 30

EXERCISES/PROBLEMS:  REQUIRED ON FINAL EXAMINATION (Monday, 5-21-12, noon to 3 pm or 1 pm to 4 pm, RM 1908, regular lecture hall) 

DISCUSSIONS WILL BE POSTED THIS WEEKEND; STAY TUNED
    
9*, 11*, 16*, 19*, 23*, 31*, 32*, 33*, 34*, 39( 39 is critical for frame work surrounding "forced" oscillations in an LRC circuit next chapter here  )
 * means reviewed in class or online.
 23. Important problem ! SEE FIGURE 30.12. After the switch is closed the magnetic energy builds up in the inductor and  heat is generated in the resistor. See equation 30.15:  I = (6/8)*(1 - e-t/T) (A)  , where T = L/R is the time constant for the circuit.

(a) Initially, dI/dt = (6/8)*(1/T)e-t/T    evaluated at t = 0.  In other words, dI/dt =
dI/dt = (6/8)*(1/T) = (6/8)*(R/L) = (6 V/2.50 H) = 2.40 A/s

(b) One brute force  way to solve this part  is to find the time when the current = 0.500 (A), then   substitute time t into the formula for dI/dt derived in part (a).
I = 0.500 = (6/8)*(1 - e-t/T) (A)  . That is:
0.500 = (0.75)*(1 - e-t/T) (A) or 
0.666 =  1 - e-t/T  
e-t/T  =  0.333
t = -(L/R)*ln(0.333).

Use your understanding of natural logs: 

Substitute (6/8)*(1/T)e-t/T    = (6/2.5)*(0.333) = 0.800 A/s.

(c) Straight  substitution: I = (6/8)*(1 - e-t/T). Note: t = 0.250 (s) and time constant T = L/R = (2.50 H/ 8.00 ohm) = 0.3125 (s). Thus: -t/T = -0.250/0.3125 = -0.8. Thus,   I =  (6/8)*(1 - e-0.800). Evaluate this to get the answer in the back of the book. 

(d)  Final steady state current occurs when  t = infinity. I = (6/8)*(1 - 0)  =
 0.75 (A).  
 31. Substitute formula Q = Qmax*cos (wt + phase) into d2Q/dt2 = - Q/(LC) .  Note: I use upper case Q to represent the charge in keeping with class notes. Qmax is the maximum or amplitude of  the oscillating charge on the capacitor plate.
32. Clearly Q = Qmax*cos wt  since the charge is a maximum at t = 0. We see that at t = 0, Q = Qmax*cos 0 = Qmax . Note the current I  as we defined in class is zero at t = 0,   since I = -dQ/dt = wQmaxsinwt. Note my negative sign in the expression I = -dQ/dt is consistent with diagrams in class where the positive current I  begins to leave the positive capacitor plate at t = 0. 
(a) w2 = 1/LC. Take the square root of both sides to find w.
(b) The electrical potential  energy at t = 0 is  UE = Qmax2/(2C). 
(c) The easy way to compute this is by substitution: I =  wQmaxsin wt. Plug in the given time t =  1.30x10 -3 (s) , get I  and finally compute UB = (1/2)*LI2. 
Another way is to use CONSERVATION OF ENERGY: You could evaluate Q at t = 1.30x10 -3 (s).  The use  Qmax2/(2C)  = Q2/(2C) +  (1/2)*LI2  to find (1/2)*LI2   and if desired  I . 
33. 
(a) w2 = 1/LC; find L. 
(b) Qmax = C*(12.0 volts).
(c) total energy = Qmax2/(2C)  = Q2/(2C) +  (1/2)*LI2  .
Thus, total energy = Qmax2/(2C)
(d) This  should be clear to you from energy considerations (see problem 32, part (c)):   total energy = Qmax2/(2C)  = Q2/(2C) +  (1/2)*LI2   .    When Q = 0, I = Imax. Thus:
Qmax2/(2C)  =  (1/2)*LImax2   .  Find the maximum current.
34. This problem is a synthesis of the previous 2 exercises. 
(a) Qmax2/(2C)  =  (1/2)*LImax2   .  Find the maximum current.
(b) Q = Qmax*cos wt . This expression is zero when wt = (2m+ 1)(pi/2), 
where pi = 3.14.... and integer m = 0, 1, 2 , 3, .... Thus t = (1/w)*(2m+ 1)(pi/2). The first zero occurs at m = 0, the second zero at m = 1. Note:  w2 = 1/LC. Also note:
At m = 0 for the first zero,  t = (1/w)*(pi/2). With w = 2*pi*f = 1/T, we get t = T/4, where T = period of oscillation. Also evaluate  at m = 1 for the second zero. 
(c) See class notes for graphs.
39, THIS PROBLEM  GIVES THE FRAMEWORK  FOR FORCED AC CIRCUITS IN CH. 31.