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Chapter 26 Part 1--Examples

Chapter 26 Part 1--Examples. Problem. If a ohmmeter is placed between points a and b in the circuits to right, what will it read?. Circuit a. The 75 and 40 W resistors are in parallel The 25 and 50 W resistors are in parallel R 75-40 =26.09 W R 50-25 =16.67 W

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Chapter 26 Part 1--Examples

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  1. Chapter 26 Part 1--Examples

  2. Problem • If a ohmmeter is placed between points a and b in the circuits to right, what will it read?

  3. Circuit a • The 75 and 40 W resistors are in parallel • The 25 and 50 W resistors are in parallel • R75-40=26.09 W • R50-25=16.67 W • Their total is 42.76 W

  4. Circuit a—Cont’d • The circled network is in parallel with the 50 W resistor so their combined resistance is 23.05 W. • This resistor is in parallel with the original 100 W resistor so the total resistance is 18.7 W

  5. Circuit b • The 60 and 20 W resistor are in parallel • The 20 W is in series with the 30 and 40 W parallel network. • R30-40 =18.0 W • R20-30-40= 38.0 W • R38-60=23.3 W

  6. Circuit b—cont’d • The 23.03 W equivalent network is in series to the 7 W resistor • This equivalent 30.03 W resistor is parallel to the 10 W resistor so • Req=7.5 W

  7. Problem • In the circuit shown, • What must be the EMF of the battery in order for a current of 2 A to flow through the 5 V battery? • How long does it take for 60 J of thermal energy to be produced in the 10 W resistor?

  8. Step 1 Reduce the Resistors • 10 + 20 =30 • (1/30)+1/60 +1/60 =4/60 so Req=15 • 1/15+1/30=3/30 so RT=10 • 10+5+5=20 • So 20 W in the upper network

  9. 20 W 5 V EMF 15 W 2 A 20 W Step 2 Reduce the EMF • -5 + 10=+5 V • So the circuit becomes:

  10. 20 W 5 V EMF 15 W 2 A 20 W Using our loop rules • -(2)*20-5-20I1=0 • I1=-2.25 A • 2=-2.25+I2 • I2=4.25 • -EMF-4.25*15+20*(-2.25) • EMF=-108.75 V • Need to reverse the battery…. I2 I1

  11. 2 Amps into the 10 W resistor • Since the equivalent resistance in the upper network is 10 W and 2 A runs through it, there is a potential difference of 20 V across each of the legs • 10+20=30 W so the current is 20/30 A=2/3 A • P=i2rso 4/9*10=40/9=4.444 W or J/s • 60=4.444 * t • t =13.5 s

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