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Transcript
Chapter 26 DC Circuits I 26-1 EMF and Terminal Voltage This resistance behaves as though it were in series with the emf. I Problem 3 3.(II) A 1.5-V dry cell can be tested by connecting it to a low-resistance ammeter. It should be able to supply at least 25 A. What is the internal resistance of the cell in this case, assuming it is much greater than that of the ammeter? I 26-1 EMF and Terminal Voltage Example 26-1: Battery with internal resistance. A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r. I 26-2 Resistors in Series and in Parallel A series connection has a single path from the battery, through each circuit element in turn, then back to the battery. I 26-2 Resistors in Series and in Parallel 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: I 26-2 Resistors in Series and in Parallel From this we get the equivalent resistance (that single resistance that gives the same current in the circuit): I 26-2 Resistors in Series and in Parallel A parallel connection splits the current; the voltage across each resistor is the same: I 26-2 Resistors in Series and in Parallel The total current is the sum of the currents across each resistor: , I 26-2 Resistors in Series and in Parallel This gives the reciprocal of the equivalent resistance: I Problem 8 8. (I) How many 10Ω resistors must be connected in series to give an equivalent resistance to five 100Ω resistors connected in parallel? I 26-2 Resistors in Series and in Parallel 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. I Circuit Analysis Principles Need to find Iin = Iout I through each element Current (charge) splits at junctions Current (charge) is the same in series V across each element or set Voltages add in series Voltages are same in parallel Voltages around a loop sum to zero DV = å å å 0 I 26-2 Resistors in Series and in Parallel Conceptual Example 26-2: Series or parallel? (a) The lightbulbs 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. I 26-2 Resistors in Series and in Parallel Example 26-4: Circuit with series and parallel resistors. How much current is drawn from the battery shown? I 26-2 Resistors in Series and in Parallel Example 26-5: Current in one branch. What is the current through the 500-Ω resistor shown? (Note: This is the same circuit as in the previous problem.) The total current in the circuit was found to be 17 mA. I 26-2 Resistors in Series and in Parallel Conceptual Example 26-6: Bulb brightness in a circuit. The circuit shown has three identical lightbulbs, each of resistance R. (a) When switch S is closed, how will the brightness of bulbs A and B compare with that of bulb C? (b) What happens when switch S is opened? Use a minimum of mathematics in your answers.