## PHYS208 Fundamentals of Physics II

### Answer for Ch. 33, 11E

0.70 ms

### Answer for Ch. 33, 18P

a) linear frequency is 275 Hz; angular frequency is 1730 rad/sec.

b) Current maximum is 365 mA.

### Answer for Ch. 33, 21P

a) *q* = *Q*/sqrt(3)

b) *t* = 0.152 T

### Answer for Ch. 33, 24P

a) *f*_{max} / *f*_{min} = 6.0

b) *C* = 36 pF in parallel; *L* = 0.22 mH.

(Note: if you get 8.6 mH you have forgotten to convert linear frequency to angular frequency!)

### Answer for Ch. 33, 25P

a) *U*_{tot} = 1.98 microJ

b) *q*_{max} = 5.56 microC

c) *i*_{max} = 12.6 mA

d) phase angle is -46.9 degrees

Why negative? We found that the cosine of the phase angle should be 0.68;
thus the phase angle may be +/- 46.9 degrees. The sign is chosen depending on
whether the capacitor is charging or discharging. To decide, look at the time derivative of the
charge at *t=0*, which is proportional to the negative of the sine of the phase angle.
For a charging capacitor, this derivative should be positive; thus the choice of -46.9 degrees.

e) phase angle is +46.9 degrees

"http://www.physics.udel.edu/~watson/phys208/exercises/answers/0424.html"

Last updated April 29, 1998.

Copyright George Watson, Univ. of Delaware, 1998.