# Chapter 18: Direct Current Circuits and Sources of EMF

This chapter discusses concepts, problems, and quick quizzes related to direct current circuits and electromotive forces, which are created by batteries, generators, thermocouples, and other means of energy conversion into electricity. The unit of EMF is Volt.

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## About Chapter 18: Direct Current Circuits and Sources of EMF

PowerPoint presentation about 'Chapter 18: Direct Current Circuits and Sources of EMF'. This presentation describes the topic on This chapter discusses concepts, problems, and quick quizzes related to direct current circuits and electromotive forces, which are created by batteries, generators, thermocouples, and other means of energy conversion into electricity. The unit of EMF is Volt.. The key topics included in this slideshow are . Download this presentation absolutely free.

## Presentation Transcript

1. Chapter 18 Chapter 18 Direct Current Circuits Direct Current Circuits Conceptual questions: 5,6,12,15,19,20 Problems: 1,8,13,40 Quick quizzes: 1,2,3

2. Sources of emf Sources of emf Electromotive forces are created by batteries, generators, thermocouples, nuclear reactors, etc. by means of chemical reactions or by conversion of kinetic, thermal, nuclear, etc. energy into electricity. Electromotive forces are created by batteries, generators, thermocouples, nuclear reactors, etc. by means of chemical reactions or by conversion of kinetic, thermal, nuclear, etc. energy into electricity. Symbol emf or Symbol emf or Unit - Volt Unit - Volt

3. emf and Internal Resistance emf and Internal Resistance A real battery has some internal resistance A real battery has some internal resistance Therefore, the terminal voltage is not equal to the emf Therefore, the terminal voltage is not equal to the emf

4. More About Internal Resistance More About Internal Resistance The schematic shows the internal resistance, r The schematic shows the internal resistance, r The terminal voltage, ΔV = V b -V a The terminal voltage, ΔV = V b -V a ΔV = ε – Ir ΔV = ε – Ir For the entire circuit, ε = IR + Ir For the entire circuit, ε = IR + Ir

5. Problem 18-1 Problem 18-1 A battery having an emf of 9.00 V delivers 117 mA when connected to a 72.0-Ω load. Determine the internal resistance of the battery. A battery having an emf of 9.00 V delivers 117 mA when connected to a 72.0-Ω load. Determine the internal resistance of the battery.

6. Resistors in Series Resistors in Series The current is the same in resistors because any charge that flows through one resistor flows through the other The current is the same in resistors because any charge that flows through one resistor flows through the other The sum of the potential differences across the resistors is equal to the total potential difference across the combination The sum of the potential differences across the resistors is equal to the total potential difference across the combination ΔV = IR 1 + IR 2 = I (R 1 +R 2 ) ΔV = IR 1 + IR 2 = I (R 1 +R 2 ) ΔV = IR eq ΔV = IR eq R eq = R 1 + R 2 + R 3 + … R eq = R 1 + R 2 + R 3 + …

7. When a piece of wire is used to connect points b and c in this figure, the brightness of bulb R 1 (a) increases, (b) decreases (c) stays the same. The brightness of bulb R 2 (a) increases, (b) decreases, (c) stays the same. QUICK QUIZ 18.1

8. QUICK QUIZ 18.2 With the switch in this circuit (figure a) closed, no current exists in R 2 because the current has an alternate zero-resistance path through the switch. Current does exist in R 1 and this current is measured with the ammeter at the right side of the circuit. If the switch is opened (figure b), current exists in R 2 . After the switch is opened, the reading on the ammeter (a) increases, (b) decreases, (c) does not change.

9. Resistors in Parallel Resistors in Parallel The potential difference across each resistor is the same because each is connected directly across the battery terminals, V = V 1 = V 2 The potential difference across each resistor is the same because each is connected directly across the battery terminals, V = V 1 = V 2 V 1 =I 1 R 1 V 2 =I 2 R 2 V 1 =I 1 R 1 V 2 =I 2 R 2 The current, I, that enters a point must be equal to the total current leaving that point The current, I, that enters a point must be equal to the total current leaving that point I = I 1 + I 2 I = I 1 + I 2 I = V 1 /R 1 + V 2 /R 2 = V (1/R 1 + 1/R 2 ) = V/R eq I = V 1 /R 1 + V 2 /R 2 = V (1/R 1 + 1/R 2 ) = V/R eq The inverse of the equivalent resistance of resistors connected in parallel is the algebraic sum of the inverses of the individual resistance The inverse of the equivalent resistance of resistors connected in parallel is the algebraic sum of the inverses of the individual resistance

10. With the switch in this circuit (figure a) open, there is no current in R 2 . There is current in R 1 and this current is measured with the ammeter at the right side of the circuit. If the switch is closed (figure b), there is current in R 2 . When the switch is closed, the reading on the ammeter (a) increases, (b) decreases, or (c) remains the same. QUICK QUIZ 18.3

11. Find the equivalent resistance of the circuit in Figure 18.8. If the total power supplied to the circuit is 4.00 W, find the emf of the battery. Problem 18.8.

12. Conceptual questions Conceptual questions 5. If you have your headlines on while you start your car why do they dim while the car starts? 5. If you have your headlines on while you start your car why do they dim while the car starts? 12. Two sets of Christmas tree lights are available. For set A, when one bulb is removed the remaining bulbs remain illuminated. For set B, when one bulb is removed the remaining bulbs do not operate. Explain the differences in wiring of the two sets. 12. Two sets of Christmas tree lights are available. For set A, when one bulb is removed the remaining bulbs remain illuminated. For set B, when one bulb is removed the remaining bulbs do not operate. Explain the differences in wiring of the two sets. Additional question: Additional question: Are the two headlights on a car wired in series or in parallel? Are the two headlights on a car wired in series or in parallel?

13. Problem 18.13. Problem 18.13. Find the current in the 12- Ω resistor Current from the battery

14. Circuits Circuits

15. Kirchhoff’s Rules Kirchhoff’s Rules Junction Rule Junction Rule The sum of the currents entering any junction must equal the sum of the currents leaving that junction The sum of the currents entering any junction must equal the sum of the currents leaving that junction Conservation of Charge Conservation of Charge Loop Rule Loop Rule The sum of the potential differences across all the elements around any closed circuit loop must be zero The sum of the potential differences across all the elements around any closed circuit loop must be zero Conservation of Energy Conservation of Energy

16. More About the Loop Rule More About the Loop Rule Traveling around the loop from a to b Traveling around the loop from a to b In a, the resistor is transversed in the direction of the current, the potential across the resistor is –IR In a, the resistor is transversed in the direction of the current, the potential across the resistor is –IR In b, the resistor is transversed in the direction opposite of the current, the potential across the resistor is +IR In b, the resistor is transversed in the direction opposite of the current, the potential across the resistor is +IR

17. Loop Rule, final Loop Rule, final In c, the source of emf is transversed in the direction of the emf (from – to +), the change in the electric potential is +ε In c, the source of emf is transversed in the direction of the emf (from – to +), the change in the electric potential is +ε In d, the source of emf is transversed in the direction opposite of the emf (from + to -), the change in the electric potential is -ε In d, the source of emf is transversed in the direction opposite of the emf (from + to -), the change in the electric potential is -ε

18. Effects of Various Currents Effects of Various Currents 5 mA or less 5 mA or less can cause a sensation of shock can cause a sensation of shock generally little or no damage generally little or no damage 10 mA 10 mA hand muscles contract hand muscles contract may be unable to let go a of live wire may be unable to let go a of live wire 100 mA 100 mA if passes through the body for 1 second or less, can be fatal if passes through the body for 1 second or less, can be fatal

19. Ground Wire Ground Wire Electrical equipment manufacturers use electrical cords that have a third wire, called a ground Electrical equipment manufacturers use electrical cords that have a third wire, called a ground Prevents shocks Prevents shocks

20. Household wiring Household wiring

21. Conceptual questions Conceptual questions 15. Embodied in Kirchhoff’s rules are two conservation laws. What are they? 15. Embodied in Kirchhoff’s rules are two conservation laws. What are they? 19. Suppose a parachutists lands on a high voltage wire and grabs a wire as she prepares to be rescued. Will she be electrocuted? If the wire breaks open, should she continue to hold onto the wire as she falls to the ground? 19. Suppose a parachutists lands on a high voltage wire and grabs a wire as she prepares to be rescued. Will she be electrocuted? If the wire breaks open, should she continue to hold onto the wire as she falls to the ground? 20. Would a fuse work successfully if it were placed in parallel with a device it was supposed to protect? 20. Would a fuse work successfully if it were placed in parallel with a device it was supposed to protect?

22. Problem 18-40 Problem 18-40 Your toaster oven and coffeemaker each dissipate 1 200 W of power. Can you operate them together if the 120-V line that feeds them has a circuit breaker rated at 15 A? Explain. Your toaster oven and coffeemaker each dissipate 1 200 W of power. Can you operate them together if the 120-V line that feeds them has a circuit breaker rated at 15 A? Explain.

23. MCAD MCAD Use the following information in questions 1-4: Use the following information in questions 1-4: Four 6-V batteries are connected in series in order to power lights A and B. The resistance of light A is 40 and the resistance of light B is 20 . Four 6-V batteries are connected in series in order to power lights A and B. The resistance of light A is 40 and the resistance of light B is 20 . 1. How does the current through light bulb A compare with the current through light bulb B? A . The current through light bulb A is less. B. The current through light bulb A is the same. C . The current through light bulb A is greater. D. None of the above is true. X X A B

24. 2. What is the potential difference between points C and D? A. 6 volts B. 12 volts C. 18 volts D. 24 volts 3. What is the current through the wire at point C? A . 0.1 A B. 0.2 A C. 0.4 A D. 1.0 A 4 . How does the voltage drop across light A compare to the drop across light B? A. The voltage drop for A is less than that for B by a factor of 4. B. The voltage drop for A is less than that for B by a factor of 2. C. The voltage drop for A is the same as that for B. D. The voltage drop for A is greater than that for B by a factor of 2. X X A B D C 40 20 6V

25. For a circuit with constant resistance, which graph represents the relation between current and potential? I V I V I V I V What must be the reading in the ammeter A for the circuit below? A. 0 A B. 6.0 A C. 8.0 A D. 12.0 A 4.0 A 6.0 A 2.0 A A A. B. C. D.