This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: PHYSZ70 Homework #3 Due in recitation week of Feb. 21 An inductor of inductance L is piaced in a circuit with a battery and two resistors as
shown below. The switch has been held open for a long time. At t=0 the switch is closed
and current begins to ﬂow through resistor R1. 1. Without doing any analysis draw a schematic curve of the current as a function of time
in resistor R1. You can determine the initial and final current based on the foliowing
points. A) Using the principle that that the current in the inductor cannot change
instantaneously, and knowing its initial value, write a formula for the current in R1 just
after the switch closes. B) After a long time currents in the resistors and inductor will
reach steady values. Knowing in that case what the voltage across the inductor must be,
you can determine the values of the final, steady currents in the two resistors and the
inductor. 2. After the above—discussed steady state is reached, the switch is opened. What will be
the voltage across the inductor just after the switch is opened? (Hint: much to the
disappointment of many of you, there wiii be no sparks because R2 is present.) 3. For the case when the switch has just been closed, label the currents in the three two—
terminal elements IL, 11,12. Write Kirchoff’s current law for the junction of R1, R2, and L.
Write Kirchoff’s voltage law for the loop including the battery, RI and R2. Relate the
voltage across the inductor to the voltage across R2. 4. (Bonus) Combine the above to write a single differential equation for the current in
the inductor. «.WMWWWWW.WMHW~MMWWWW “TZéneﬂ a“ CuVMn‘} f/augwmﬁggﬁh i
E menm. . r“: r. "f Tm. i
% E E 3 :
mmii'zr"; i 4% F: W s "r E“ uwmﬁ V “a€1w$+g_j§¥ﬂm(ggial m M, mi mm“ M. A w m WWWWW , W m m M WWWW km-m-,m~mm.mmmmWWWWW.
n W “diam Icy? WW ya 3" Png—Hzmtlp1 ZMYDW M “W H W W- .W M w W
WM L M W m M W W W W W M Wm. mm M MW mm WW WWW m m [VDLU Q LIMINME 5,31 = Va. 1/ Q amumwnmm ...
View Full Document
- Fall '08