solutions_chapter19

# Set up where i is the current through the battery and

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Set Up: where I is the current through the battery and V is the potential difference applied to the person. Solve: This is well within the range of common household voltages. 19.22. Set Up: Current is measured in A, potential difference in V. Problem 19.21 says the resistance of a wet human is 1 k . Solve: (a) Volts are not current; he probably meant a potential difference of 12 V. (b) Problem 19.21 says currents above 10 mA are dangerous, so this current is dangerous. 19.23. Set Up: and Solve: 19.24. Set Up: Table 19.1 gives for gold. Solve: (a) (b) 19.25. Set Up: so Solve: (a) (b) I new 5 V p d new 2 4 r L 5 V p 1 2 d 2 2 4 r L 5 4 I d new 5 2 d . I new 5 V p d 2 4 r L new 5 V p d 2 4 r 1 2 L 2 5 1 2 I L new 5 2 L . I 5 V R 5 VA r L 5 V p 1 d / 2 2 2 r L 5 V p d 2 4 r L . V 5 IR R 5 r L A . V 5 IR 5 1 1.15 A 21 0.282 V 2 5 0.324 V R 5 4 r L p d 2 5 4 1 2.44 3 10 2 8 V # m 21 6.40 m 2 p 1 0.840 3 10 2 3 m 2 2 5 0.282 V r 5 2.44 3 10 2 8 V # m V 5 IR . R 5 r L A 5 r L p 1 d / 2 2 2 . 1 V 5 1 V / A 5 1 kg # m 2 / A 2 # s 3 1 V 5 1 N # m / C 5 1 1 kg # m / s 2 2 # m / C 5 1 1 kg # m 2 / s 2 2 / A # s 5 1 kg # m 2 / A # s 3 1 N 5 1 kg # m / s 2 1 V 5 1 N # m / C. 1 V 5 1 V / A I 5 V R 5 6 V 1 k V 5 16 mA. V 5 IR . V V 5 IR 5 1 5.0 3 10 2 3 A 21 1000 V 2 5 5.0 V. V 5 IR , E 5 IR 5 1 0.453 A 21 83.3 V 2 5 37.7 V I 5 E R 5 12.0 V 83.3 V 5 0.144 A R 5 E I 5 1.50 V 18.0 3 10 2 3 A 5 83.3 V E 5 IR . 4.80 4.40 4.00 3.60 3.20 2.80 2.40 0.00 1.00 2.00 3.00 4.00 V ab 1 V 2 ( a ) I 1 A 2 I 1 A 2 R 1 Ohms 2 ( b ) 6.00 5.00 4.00 3.00 2.00 1.00 0.00 1.00 2.00 3.00 4.00 Thyrite TM V ab R 5 V ab I 19-4 Chapter 19

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(c) Reflect: I increases when R decreases. R decreases when L decreases or d increases. 19.26. Set Up: so and 19.27. Set Up: When the switch is open there is no current and the terminal voltage of the battery equals its emf, When the switch is closed, current I flows and the terminal voltage V of the battery is The current I is the same at all points of the circuit. Solve: with and so and Reflect: When current flows through the battery, the terminal voltage is less than the emf because of the voltage across the internal resistance. 19.28. Set Up: When the switch is open, there is no current. The voltmeter reads the voltage across the circuit element it is connected across. The emf of the battery is 12.0 V. Solve: (a) so the terminal voltage of the battery equals the emf. The voltmeter reads 12.0 V. (b) so the voltage across the resistor is zero. The voltmeter reads zero. (c) The sum of the potential changes around the circuit must be zero, so if there is a potential rise of 12.0 V across the battery, there must be a potential drop of 12.0 V across the switch. The voltmeter reads 12.0 V. (d) and (a) The terminal voltage of the battery is (b) The voltage across the resistor is (c) The meter is now a path of zero resistance and the voltage across it is zero. 19.29. Set Up: The terminal voltage of the battery is given by The terminal voltage also equals the voltage across the resistor, so We have two unknowns so we need two equations.
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