Chapter 26:DC Circuits Chapter 26 Opener. These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram.

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Chapter 26:DC Circuits Chapter 26 Opener. These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram below shows a possible amplifier circuit for each stereo channel. Although the large triangle is an amplifier chip containing transistors (discussed in Chapter 40), the other circuit elements are ones we have met, resistors and capacitors, and we discuss them in circuits in this Chapter. We also discuss voltmeters and ammeters, and how they are built and used to make measurements. HW4: Chapter 25: Pb. 19, Pb.25, Pb. 31; Chapter 26: Pb 18, Pb.32, Pb.50, Pb. 51 Due Wednesday, March 8

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? (c) What is the current in the 6.0-Ω resistor? Solution: a. First, find the equivalent resistance. The 8-Ω and 4-Ω resistors in parallel have an equivalent resistance of 2.7 Ω; this is in series with the 6.0-Ω resistor, giving an equivalent of 8.7 Ω. This is in parallel with the 10.0-Ω resistor, giving an equivalent of 4.8 Ω; finally, this is in series with the 5.0-Ω resistor and the internal resistance of the battery, for an overall resistance of 10.3 Ω. The current is 9.0 V/10.3 Ω = 0.87 A. b. The terminal voltage of the battery is the emf less Ir = 8.6 V. c. The current across the 6.0-Ω resistor and the 2.7-Ω equivalent resistance is the same; the potential drop across the 10.0-Ω resistor and the 8.7-Ω equivalent resistor is also the same. The sum of the potential drop across the 8.7-Ω equivalent resistor plus the drops across the 5.0-Ω resistor and the internal resistance equals the emf of the battery; the current through the 8.7-Ω equivalent resistor is then 0.48 A.

26-3 Kirchhoff’s Rules Some circuits cannot be broken down into series and parallel connections. For these circuits we use Kirchhoff’s rules. Figure 26-11. Currents can be calculated using Kirchhoff’s rules.

26-3 Kirchhoff’s Rules Junction rule: The sum of currents entering a junction equals the sum of the currents leaving it.

26-3 Kirchhoff’s Rules Loop rule: The sum of the changes in potential around a closed loop is zero. Figure 26-12. Changes in potential around the circuit in (a) are plotted in (b).

Loop Rule Traveling around the loop from a to b In (a), the resistor is traversed in the direction of the current, the potential across the resistor is –IR In (b), the resistor is traversed in the direction opposite of the current, the potential across the resistor is +IR

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

Circuit Analysis Principles Keep in mind that: I through each element Current (charge) splits at junctions Current (charge) is the same in series Apply junction rule V across each element or set Voltages add in series Voltages are same in parallel Voltages around a loop sum to zero (loop rule)

26-3 Kirchhoff’s Rules Example 26-9: Using Kirchhoff’s rules. Calculate the currents I1, I2, and I3 in the three branches of the circuit in the figure. Solution: You will have two loop rules and one junction rule (there are two junctions but they both give the same rule, and only 2 of the 3 possible loop equations are independent). Algebraic manipulation will give I1 = -0.87 A, I2 = 2.6 A, and I3 = 1.7 A.

26-3 Kirchhoff’s Rules Problem Solving: Kirchhoff’s Rules Label each current, including its direction. Identify unknowns. Apply junction and loop rules; you will need as many independent equations as there are unknowns. 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.

Problem 31 31. (II) (a) What is the potential difference between points a and d in Fig. 26–49 (similar to Fig. 26–13, Example 26–9), and (b) what is the terminal voltage of each battery?