1. Use CONS. of POWER.  Ignore the CAP.
(30 V)(2 mA) means 60 mW is output from
EMFA.  (5000 Ohms)(2 mA)^2 means 20 mW
appears in RES.  So 40 mW is being stored
in CAP.

2. Value is 1/3 of final value in 30 micros,
so exp(-30micros/tau)=2/3, so tau=30micros/0.405
or 74.0 micros.  The time constant for LR circuits
is L/R, so since R=1000 Ohms, then L is 74.0 mH.

3. The 0.5 A current in the battery means the
size of the inductor voltage is 12 V-10 V=2 V.
So dI/dt = 2 V/10 mH = 200 A/s.

4. For LC, the ang. freq. is sqrt(1/LC), so
this is a straightforward substition, yielding
16.7 rad/s.

5. Just after, there is no current, thus
no resistor voltage, so the inductor voltage
is 240 V, with the a side high.

6. Like #3, a straightforward substitution for
omega in rad/s, then use units to convert to period
in s/cycle.  2pi rad/cycle/(16.7 rad/s) yields
0.376 s/cycle.

7. AC power is Irms*Vrms so the rms current is
10 A; then since Vrms=Irms*R, R must be 11 Ohms.

8. Use impedance of LRC series with generator,
with L set to zero.  The capacitive reactance
is 1/((377 rad/s)(2 microF) so 1326 Ohms;
added "quadratically" to the 2000 Ohms resistance
yields 2400 Ohms for the impedance.

9. Use impedance of LRC series with generator,
with C set to infinity, i.e. X_C set to 0,
and R set to zero, since the resistance of an
ammeter is tiny.  The inductive reactance is
(377 rad/s)(0.5 H) or 189 Ohms, so this is the
impedance, and the rms I is rms V/Z or 0.584 A.

10. Like #3, a straightforward sub for omega
in rad/s, then use units to convert to freq.
in cycles/s.  (10000 rad/s)/(2pi rad/cycle)
yields 1592 Hz.

11. Use impedance of LRC series with generator,
in a straightforward substitution.  The ang.
freq. is 500 rad/s so inductive reactance is
500 Ohms and capacitive reactance is
(1/.0005) Ohms = 2000 Ohms.  So the net
reactance is 1500 Ohms and the resistance is
1000 Ohms, so "adding quadratically" yields
an impedance of 1803 Ohms.

12. Like #9, a straightforward sub for omega
in rad/s, followed by a unit conversion to
cycles/s.  (1000 rad/s)/(2pi rad/cycle)
yields 159.2 Hz.

13. First, #10 must be done to find the
impedance of 1803 Ohms; then, Irms=Vrms/Z
were Vrms=140/sqrt(2) which is 99.0 V.
So Irms=54.9 mA and, for the capacitor,
the rms capacitor voltage is IrmsX_C
or 110 V.

14. For such an AC, the power loss is
Irms^2R so and Irms=3 A/sqrt(2) so the
power loss is (4.5 A^2)(100 Ohms)=450 W.

15. Use impedance of LRC series with generator,
with C set to infinity, i.e. X_C set to 0.
The inductive reactance is (377 rad/s)(0.01 H)
for 3.77 Ohms.  "Adding quadratically" to
the 10 Ohms resistance yields an impedance
of 10.68 Ohms, so the peak current is
110 V/10.68 Ohms for 10.3 A.

16. A straightforward use of Erad=c*Brad.
(600 N/C)/(3e8 m/s) = 2e-6 N/C/(m/s)
or 2 microT.

17. Use spherical symmetry and the area
of a sphere, i.e. 4pir^2; so the energy
passing per second per square meter is
(100 W)/(4pi*1m^2)=7.96 W/m^2.  (GBA notes
that asking for "the magnitude of the
Poynting vector" is equivalent to asking
for the intensity of the light; we will
not use the Poynting vector language in
our class.)

18. Erad and Brad are always perpendicular
to one another. The direction of the
energy propagation is the direction of
the vector EradcrossBrad; the mathematical
expression of this idea has vector c divided
by magnitude c, etc.

19. The energy density in a static E field
is 0.5(epsilon0)E^2.  In an EM wave, there
is energy in both the Erad and the Brad,
but because Erad=cBrad, these two energy
densities are equally large.  So at any
place, the time-dependent energy density in
the EM wave is 2*0.5*(epsilon0)Erad,max^2(t).
Since the time dependence is sinusoidal,
when the energy density is averaged over
time, another factor of 0.5 emerges, so that
the time-average energy density is just
0.5(epsilon0)Erad,max^2.

20. The displacement current through the
ENTIRE capacitor is no different from the
current into, or out of, the capacitor.
The current in is given as 12 nC/s, so
that is also the displacement current of
1.2e-8 C/s or A.