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Copyright © 2009 Pearson Education, Inc. Chapter 26 DC Circuits

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Copyright © 2009 Pearson Education, Inc. A series connection has a single path from the battery, through each circuit element in turn, then back to the battery. 26-2 Resistors in Series and in Parallel

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Copyright © 2009 Pearson Education, Inc. The current through each resistor is the same; the voltage depends on the resistance. The sum of the voltage drops across the resistors equals the battery voltage: 26-2 Resistors in Series

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Copyright © 2009 Pearson Education, Inc. From this we get the equivalent resistance (that single resistance that gives the same current in the circuit): 26-2 Resistors in Series Unless an internal resistance r is specified assume V constant.

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ConcepTest 26.1aSeries Resistors I 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R

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equal evenly 3 V drop Since the resistors are all equal, the voltage will drop evenly across the 3 resistors, with 1/3 of 9 V across each one. So we get a 3 V drop across each. ConcepTest 26.1aSeries Resistors I 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R

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ConcepTest 26.1bSeries Resistors II 12 V R 1 = 4 R 2 = 2 In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V

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ConcepTest 26.1bSeries Resistors II 12 V R 1 = 4 R 2 = 2 In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V The voltage drop across R 1 has to be twice as big as the drop across R 2.V 1 = 8 V The voltage drop across R 1 has to be twice as big as the drop across R 2. This means that V 1 = 8 V and V 2 = 4 V. Or else you could find the current I = V/R = (12 V)/(6 = 2 A, and then use Ohm’s law to get voltages. Follow-up: What happens if the voltage is 24 V?

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Copyright © 2009 Pearson Education, Inc. A parallel connection splits the current; the voltage across each resistor is the same: 26-2 Resistors in Parallel

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Copyright © 2009 Pearson Education, Inc. The total current is the sum of the currents across each resistor: 26-2 Resistors in Parallel,

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Copyright © 2009 Pearson Education, Inc. An analogy using water may be helpful in visualizing parallel circuits. The water (current) splits into two streams; each falls the same height, and the total current is the sum of the two currents. With two pipes open, the resistance to water flow is half what it is with one pipe open. 26-2 Resistors in Parallel

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ConcepTest 26.2aParallel Resistors I In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5 R 2 = 2 1) 10 A 2) zero 3) 5 A 4) 2 A 5) 7 A

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voltagesame V 1 = I 1 R 1 I 1 = 2 A The voltage is the same (10 V) across each resistor because they are in parallel. Thus, we can use Ohm’s law, V 1 = I 1 R 1 to find the current I 1 = 2 A. ConcepTest 26.2aParallel Resistors I In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5 R 2 = 2 1) 10 A 2) zero 3) 5 A 4) 2 A 5) 7 A Follow-up: What is the total current through the battery?

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ConcepTest 26.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit?

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ConcepTest 26.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero resistance of the circuit drops resistance decreasescurrent must increase As we add parallel resistors, the overall resistance of the circuit drops. Since V = IR, and V is held constant by the battery, when resistance decreases, the current must increase. Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? Follow-up: What happens to the current through each resistor?

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Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Conceptual Example 26-2: Series or parallel? (a) The light bulbs in the figure are identical. Which configuration produces more light? (b) Which way do you think the headlights of a car are wired? Ignore change of filament resistance R with current. Note: brightness is proportional to power

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Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Conceptual Example 26-2: Series or parallel? (a) The light bulbs in the figure are identical. Which configuration produces more light? (b) Which way do you think the headlights of a car are wired? Ignore change of filament resistance R with current.

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Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Conceptual Example 26-3: An illuminating surprise. A 100-W, 120-V lightbulb and a 60-W, 120-V lightbulb are connected in two different ways as shown. In each case, which bulb glows more brightly? Ignore change of filament resistance with current (and temperature).

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Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Conceptual Example 26-3: An illuminating surprise.

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Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Example: Current in one branch. What is the current through the 500-Ω resistor shown?

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Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Example 26-8: Analyzing a circuit. A 9.0-V battery whose internal resistance r is 0.50 Ω is connected in the circuit shown. (a) How much current is drawn from the battery? (b) What is the terminal voltage of the battery? Note: slight error in figure and text

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Copyright © 2009 Pearson Education, Inc. Some circuits cannot be broken down into series and parallel connections. For these circuits we use Kirchhoff’s rules. 26-3 Kirchhoff’s Rules

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Copyright © 2009 Pearson Education, Inc. Junction rule: The sum of currents entering a junction equals the sum of the currents leaving it. 26-3 Kirchhoff’s Rules

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Copyright © 2009 Pearson Education, Inc. Loop rule: The sum of the changes in potential around a closed loop is zero. 26-3 Kirchhoff’s Rules

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Copyright © 2009 Pearson Education, Inc. Problem Solving: Kirchhoff’s Rules 1. Label each current, including its direction. 2. Identify unknowns. 3. Apply junction and loop rules; you will need as many independent equations as there are unknowns. Each new junction or loop must include a new element. 4.Solve the equations, being careful with signs. If the solution for a current is negative, that current is in the opposite direction from the one you have chosen. 26-3 Kirchhoff’s Rules

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ConcepTest 26.10Kirchhoff’s Rules ConcepTest 26.10 Kirchhoff’s Rules The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes

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the point between the bulbs is at 12 VBut so is the point between the batteries When the switch is open, the point between the bulbs is at 12 V. But so is the point between the batteries. If there is no potential difference, then no current will flow once the switch is closed!! Thus, nothing changes. The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes ConcepTest 26.10Kirchhoff’s Rules ConcepTest 26.10 Kirchhoff’s Rules 24 V Follow-up: Follow-up: What happens if the bottom battery is replaced by a 24 V battery?

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Copyright © 2009 Pearson Education, Inc. 26-3 Kirchhoff’s Rules Example: Using Kirchhoff’s rules. Calculate the currents I 1, I 2, and I 3 in the three branches of the circuit in the figure.

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Copyright © 2009 Pearson Education, Inc. Solution:

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ConcepTest 26.12Kirchhoff’s Rules ConcepTest 26.12 More Kirchhoff’s Rules 2 V 2 2 V 6 V 4 V 3 1 I1I1 I3I3 I2I2 Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I 2 + 6 = 0 5) 2 – I 1 – 3I 3 – 6 = 0

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ConcepTest 26.12Kirchhoff’s Rules ConcepTest 26.12 More Kirchhoff’s Rules 2 V 2 2 V 6 V 4 V 3 1 I1I1 I3I3 I2I2 Eq. 3 is valid for the left loop Eq. 3 is valid for the left loop: The left battery gives +2 V, then there is a drop through a 1 resistor with current I 1 flowing. Then we go through the middle battery (but from + to – !), which gives –4 V. Finally, there is a drop through a 2 resistor with current I 2. Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I 2 + 6 = 0 5) 2 – I 1 – 3I 3 – 6 = 0

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Copyright © 2009 Pearson Education, Inc. EMFs in parallel only make sense if the voltages are the same; this arrangement can produce more current than a single emf. 26-4 Series and Parallel EMFs

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Copyright © 2009 Pearson Education, Inc. When the switch is closed, the capacitor will begin to charge. As it does, the voltage across it increases, and the current through the resistor decreases. 26-5 Circuits Containing Resistor and Capacitor (RC Circuits)

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Copyright © 2009 Pearson Education, Inc. 26-5 RC Circuits To find the voltage as a function of time, we write the equation for the voltage changes around the loop: Since Q = dI/dt, we can integrate to find the charge as a function of time:

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Copyright © 2009 Pearson Education, Inc. 26-5 RC Circuits The voltage across the capacitor is V C = Q/C : The quantity RC that appears in the exponent is called the time constant of the circuit:

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Copyright © 2009 Pearson Education, Inc. 26-5 RC Circuits The current at any time t can be found by differentiating the charge:

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Copyright © 2009 Pearson Education, Inc. 26-5 RC Circuits Example 26-11: RC circuit, with emf. The capacitance in the circuit shown is C = 0.30 μ F, the total resistance is 20 kΩ, and the battery emf is 12 V. Determine (a) the time constant, (b) the maximum charge the capacitor could acquire, (c) the time it takes for the charge to reach 99% of this value, (d) the current I when the charge Q is half its maximum value, (e) the maximum current, and (f) the charge Q when the current I is 0.20 its maximum value.

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Copyright © 2009 Pearson Education, Inc. If an isolated charged capacitor is connected across a resistor, it discharges: 26-5 RC Circuits

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Copyright © 2009 Pearson Education, Inc. 26-5 RC Circuits Once again, the voltage and current as a function of time can be found from the charge: and

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Copyright © 2009 Pearson Education, Inc. 26-5 RC Circuits Example 26-12: Discharging RC circuit. In the RC circuit shown, the battery has fully charged the capacitor, so Q 0 = C E. Then at t = 0 the switch is thrown from position a to b. The battery emf is 20.0 V, and the capacitance C = 1.02 μF. The current I is observed to decrease to 0.50 of its initial value in 40 μs. (a) What is the value of Q, the charge on the capacitor, at t = 0? (b) What is the value of R ? (c) What is Q at t = 60 μs?

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Copyright © 2009 Pearson Education, Inc. 26-5 RC Circuits Conceptual Example 26-13: Bulb in RC circuit. In the circuit shown, the capacitor is originally uncharged. Describe the behavior of the lightbulb from the instant switch S is closed until a long time later.

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Copyright © 2009 Pearson Education, Inc. Most people can “feel” a current of 1 mA; a few mA of current begins to be painful. Currents above 10 mA may cause uncontrollable muscle contractions, making rescue difficult. Currents around 100 mA passing through the torso can cause death by ventricular fibrillation. Higher currents may not cause fibrillation, but can cause severe burns. Household voltage can be lethal if you are wet and in good contact with the ground. Be careful! 26-6 Electric Hazards

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Copyright © 2009 Pearson Education, Inc. A source of emf transforms energy from some other form to electrical energy. A battery is a source of emf in parallel with an internal resistance. Resistors in series: Summary of Chapter 26

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Copyright © 2009 Pearson Education, Inc. Resistors in parallel: Kirchhoff’s rules: 1.Sum of currents entering a junction equals sum of currents leaving it. 2.Total potential difference around closed loop is zero. Summary of Chapter 26

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Copyright © 2009 Pearson Education, Inc. RC circuit has a characteristic time constant: To avoid shocks, don’t allow your body to become part of a complete circuit. Summary of Chapter 26

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