Solutions to Homework Questions 7
Solutions to Homework Questions 7
Solutions to Homework Questions 7
- No tags were found...
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
Physics 112 <strong>Homework</strong> 7 (solutions) (2004 Fall)<br />
Chapt21, Problem-33: An RLC circuit is used <strong>to</strong> tune a radio <strong>to</strong> an FM station broadcasting at 88.9 MHz.<br />
The resistance in the circuit is 12.0 ! and the capacitance is 1.40 pF. What inductance should be present in the circuit<br />
Solution:<br />
The resonance frequency of the circuit should match the broadcast frequency of the station.<br />
1<br />
f 0 =<br />
2! LC gives L = 1<br />
4! 2 f 2 0<br />
C ,<br />
1<br />
or L =<br />
2 .29 µH<br />
4! 2 88.9 "10 6 Hz<br />
( ) 2 ( 1.40 " 10 #12 F) = 2 .29" 10#6 H =<br />
Chapt21, Problem-35: The AM band extends from approximately 500 kHz <strong>to</strong> 1 600 kHz. If a<br />
2.0-µH induc<strong>to</strong>r is used in a tuning circuit for a radio, what are the extremes that a capaci<strong>to</strong>r must reach in<br />
order <strong>to</strong> cover the complete band of frequencies<br />
Solution:<br />
1<br />
f 0 =<br />
2! LC , so C = 1<br />
4! 2 f 2 0<br />
L<br />
For f 0 = f 0<br />
For f 0 = f 0<br />
( ) min<br />
= 500 kHz = 5.00 !10 5 Hz ,<br />
C = C max =<br />
1<br />
( ) 2 ( 2.0 " 10 #6 H) = 5.1" 10#8 F =<br />
4! 2 5.00 "10 5 Hz<br />
( ) max<br />
= 1600 kHz = 1.60! 10 6 Hz ,<br />
C = C min =<br />
1<br />
( ) 2 ( 2.0 " 10 #6 H) = 4.9 " 10#9 F =<br />
4! 2 1.60 " 10 6 Hz<br />
51 nF<br />
4.9 nF<br />
Chapt21, Problem-41: A transformer on a pole near a fac<strong>to</strong>ry steps the voltage down from 3600 V <strong>to</strong> 120V.<br />
The transformer is <strong>to</strong> deliver 1 000 kW <strong>to</strong> the fac<strong>to</strong>ry at 90% efficiency. Find (a) the power delivered <strong>to</strong> the primary, (b) the<br />
current in the primary, and (c) the current in the secondary.<br />
Solution:<br />
(a) At 90% efficiency, (! av ) output<br />
= 0.90 (! av ) input<br />
.<br />
Thus, if (! av ) output<br />
= 1000 kW ,<br />
(<br />
the input power <strong>to</strong> the primary is (! av ) input<br />
= ! av ) output 1000 kW<br />
= =<br />
0.90 0.90<br />
1.1! 103 kW<br />
(<br />
(b) I 1, rms = ! av ) input<br />
= 1.1 #103 kW<br />
= 1.1# 106 W<br />
"V 1, rms<br />
"V 1, rms<br />
3600 V = 3.1! 102 A<br />
(<br />
(c) I 2, rms = ! av ) output 1000 kW<br />
= = 1.0 #106 W<br />
"V 2, rms<br />
"V 1, rms<br />
120 V = 8.3 !103 A<br />
3