Chapter 26 Part 1--ExamplesProblemCircuit aCircuit a—Cont’dCircuit bCircuit b—cont’dSlide 7Step 1 Reduce the ResistorsStep 2 Reduce the EMFUsing our loop rules2 Amps into the 10 W resistor1Chapter 26 Part 1--Examples2ProblemIf a ohmmeter is placed between points a and b in the circuits to right, what will it read?3Circuit aThe 75 and 40 resistors are in parallelThe 25 and 50 resistors are in parallelR75-40=26.09 R50-25=16.67 Their total is 42.76 4Circuit a—Cont’dThe circled network is in parallel with the 50 resistor so their combined resistance is 23.05 .This resistor is in parallel with the original 100 resistor so the total resistance is 18.7 5Circuit bThe 60 and 20 resistor are in parallelThe 20 is in series with the 30 and 40 parallel network.R30-40 =18.0 R20-30-40= 38.0 R38-60=23.3 6Circuit b—cont’dThe 23.03 equivalent network is in series to the 7 resistorThis equivalent 30.03 resistor is parallel to the 10 resistor soReq=7.5 7ProblemIn the circuit shown, 1. What must be the EMF of the battery in order for a current of 2 A to flow through the 5 V battery?2. How long does it take for 60 J of thermal energy to be produced in the 10 resistor?8Step 1 Reduce the Resistors10 + 20 =30(1/30)+1/60 +1/60 =4/60 so Req=151/15+1/30=3/30 so RT=1010+5+5=20So 20 in the upper network9Step 2 Reduce the EMF-5 + 10=+5 VSo the circuit becomes:20 5 VEMF15 20 2 A10Using our loop rules-(2)*20-5-20I1=0I1=-2.25 A2=-2.25+I2I2=4.25-EMF-4.25*15+20*(-2.25)EMF=-108.75 VNeed to reverse the battery….20 5 VEMF15 20 2 AI1I2112 Amps into the 10 resistorSince the equivalent resistance in the upper network is 10 and 2 A runs through it, there is a potential difference of 20 V across each of the legs10+20=30 so the current is 20/30 A=2/3 AP=i2r so 4/9*10=40/9=4.444 W or J/s 60=4.444 * tt =13.5
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