ch31 - CHAPTER Alternating-Current Circuits 31 Note: Unless...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 31 Alternating-Current Circuits Note: Unless otherwise indicated, the symbols I , V , E , and P denote the rms values of I , V , and E and the average power. 1* · A 200-turn coil has an area of 4 cm 2 and rotates in a magnetic field of 0.5 T. ( a ) What frequency will generate a maximum emf of 10 V? ( b ) If the coil rotates at 60 Hz, what is the maximum emf? ( a ) E = NBA ? cos ? t (see Problem 30-8-5) ( b ) E max = NBA ? = 2 p NBAf ? = E max / NBA = 250 s –1 ; f = ? /2 p = 39.8 Hz E max = 15.1 V 2 · In what magnetic field must the coil of Problem 1 be rotating to generate a maximum emf of 10 V at 60 Hz? Use Equ. 31–4; solve for B B = 0.332 T 3 · A 2-cm by 1.5-cm rectangular coil has 300 turns and rotates in a magnetic field of 4000 G. ( a ) What is the maximum emf generated when the coil rotates at 60 Hz? ( b ) What must its frequency be to generate a maximum emf of 110 V? ( a ) Use Equ. 31-4 ( b ) Use Equ. 31-4; solve for f = ? /2 p E max = 13.6 V f = 486 Hz 4 · The coil of Problem 3 rotates at 60 Hz in a magnetic field B . What value of B will generate a maximum emf of 24 V? Use Equ. 31-4; solve for B B = 0.707 T 5* · As the frequency in the simple ac circuit in Figure 31-26 increases, the rms current through the resistor ( a ) increases. ( b ) does not change. ( c ) may increase or decrease depending on the magnitude of the original frequency. ( d ) may increase or decrease depending on the magnitude of the resistance. ( e ) decreases. ( b ) 6 · If the rms voltage in an ac circuit is doubled, the peak voltage is ( a ) increased by a factor of 2. ( b ) decreased by a factor of 2. ( c ) increased by a factor of 2 . ( d ) decreased by a factor of 2 . ( e ) not changed. ( a ) 7 · A 100-W light bulb is plugged into a standard 120-V (rms) outlet. Find ( a ) I rms , ( b ) I max , and ( c ) the maximum power. ( a ) Use Equ. 31-14 ( b ) Use Equ. 31-12 I rms = 0.833 A I max = 1.18 A
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 31 Alternating-Current Circuits ( c ) P max = I max E max = 2 I rms E rms = 2 P av P max = 200 W 8 · A 3- O resistor is placed across a generator having a frequency of 60 Hz and a maximum emf of 12.0 V. ( a ) What is the angular frequency ? of the current? ( b ) Find I max and I rms . What is ( c ) the maximum power into the resistor, ( d ) the minimum power, and ( e ) the average power? ( a ) ? = 2 p f ( b ) Use Equs. 31-8 and 31-12 ( c ) P max = I max 2 R ( d ) P min = ( I min ) 2 R ( e ) P av = 1/2 P max ? = 377 rad/s I max = 4 A; I rms = 2.83 A P max = 48 W P min = 0 P av = 24 W 9* · A circuit breaker is rated for a current of 15 A rms at a voltage of 120 V rms. ( a ) What is the largest value of I max that the breaker can carry? ( b ) What average power can be supplied by this circuit? ( a
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/01/2008 for the course PHYS 13/14 taught by Professor Millan/thorstensen during the Spring '08 term at Dartmouth.

Page1 / 26

ch31 - CHAPTER Alternating-Current Circuits 31 Note: Unless...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online