1. 1 Chapter 26 Part 1--Examples

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

3. 3 Circuit a The 75 and 40  resistors are in parallel The 25 and 50  resistors are in parallel R 75-40 =26.09  R 50-25 =16.67  Their total is 42.76 

4. 4 Circuit a—Cont’d The 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 

5. 5 Circuit b The 60 and 20  resistor are in parallel The 20  is in series with the 30 and 40  parallel network. R 30-40 =18.0  R 20-30-40 = 38.0  R 38-60 =23.3 

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

7. 7 Problem In 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?

8. 8 Step 1 Reduce the Resistors 10 + 20 =30 (1/30)+1/60 +1/60 =4/60 so R eq =15 1/15+1/30=3/30 so R T =10 10+5+5=20 So 20  in the upper network

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

10. 10 Using our loop rules -(2)*20-5-20I 1 =0 I 1 =-2.25 A 2=-2.25+I 2 I 2 =4.25 -EMF-4.25*15+20*(- 2.25) EMF=-108.75 V Need to reverse the battery…. 20  5 V EMF 15  20  2 A I1I1 I2I2

11. 11 2 Amps into the 10  resistor Since 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 legs 10+20=30  so the current is 20/30 A=2/3 A P=i 2 r so 4/9*10=40/9=4.444 W or J/s 60=4.444 * t t =13.5 s