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Chapter 26 Part 1--Examples 1 Problem If a ohmmeter is placed between points a and b in the circuits to right, what will it read? 2 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 3 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 4 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 5 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 6 Problem In the circuit shown, 1. 2. 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? 7 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 8 Step 2 Reduce the EMF -5 + 10=+5 V So the circuit becomes: 20 W 2A 5V 15 W EMF 20 W 9 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…. 20 W 2A 5V 15 W EMF I2 20 W I1 10 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=i2r so 4/9*10=40/9=4.444 W or J/s 60=4.444 * t t =13.5 s 11