# Reflect the power output of the 12 v battery is less

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Reflect: The power output of the 12 V battery is less than because of energy dissipated in the internal resistance of the battery. When current flows through a battery from the positive to the negative terminal, the positive charge E I I 2 R total 5 1 0.50 A 2 2 1 16 V 2 5 4.0 W. 6.0 W 2 2.0 W 5 4.0 W. 1 4.0 V 21 0.50 A 2 5 2.0 W 1 12.0 V 21 0.50 A 2 5 6.0 W P 5 1 4 V 21 0.50 A 2 1 1 0.50 A 2 2 1 4 V 2 5 2.0 W 1 1.0 W 5 3.0 W I 2 r . E I P 5 E I 2 I 2 r 5 1 12.0 V 21 0.50 A 2 2 1 0.50 A 2 2 1 2 V 2 5 6.0 W 2 0.5 W 5 5.5 W. P 5 I 2 1 3.00 V 1 7.00 V 2 5 1 0.50 A 2 2 1 10.0 V 2 5 2.5 W. I 5 8 V 16 V 5 0.50 A. 1 12 V 2 4 V 2 I 1 2 V 1 3 V 1 4 V 1 7 V 2 5 0. P E 5 E I . P R 5 I 2 R . V ab 5 16.0 V 2 1 0.471 A 21 1.6 V 2 5 15.2 V V b 1 16.0 V 2 I 1 1.6 V 2 5 V a . I 5 16.0 V 2 8.0 V 17.0 V 5 0.471 A. 1 16.0 V 2 I 1 1.6 V 1 5.0 V 1 1.4 V 1 9.0 V 2 2 8.0 V 5 0. q t (a) Q max i t (b) I max t i I max t q Q max 19-18 Chapter 19
passing through the battery in the direction of the current loses electrical energy. In the battery, electrical energy is removed from the circuit. The power in this case corresponds to the rate at which chemical energy is being stored. 19.74. Set Up: In 1.00 s the electron passes a point on the orbit times. The charge of an electron has magnitude e. Solve: The magnitude of the average current is The direction of the current is opposite to the direction of circulation of the electron, since the electron has negative charge. 19.75. Set Up: When the temperature increases the resistance increases and the current decreases. Solve: (a) (b) (i) (ii) 19.76. Set Up: so each piece has resistance For resistors in parallel, Solve: This can also be obtained as decreasing L by and increasing A by 3, for an overall decrease of 19.77. Set Up: The circuit is sketched in Figure 19.77. is the combined internal resistance of both batteries. The power delivered to the bulb is Figure 19.77 Solve: (a) The loop rule gives This is also (b) (c) so The loop rule gives Reflect: When the power to the bulb has decreased to half its initial value, the total internal resistance of the two bat- teries is nearly half the resistance of the bulb. Compared to a single battery, using two identical batteries in series doubles the emf but also doubles the total internal resistance. r total 5 3.0 V 2 1 0.125 A 21 17 V 2 0.125 A 5 7.0 V . 1.5 V 1 1.5 V 2 IR 2 Ir total . I 5 Å P R 5 Å 0.265 W 17 V 5 0.125 A. P 5 I 2 R P 5 0.530 W 2 5 0.265 W. Energy 5 1 0.530 W 21 5.0 h 21 3600 s / h 2 5 9540 J 1 3.0 V 21 0.1765 A 2 . P 5 I 2 R 5 1 0.1765 A 2 2 1 17 V 2 5 0.530 W. I 5 0.1765 A. 1.5 V 1 1.5 V 2 I 1 17 V 2 5 0. r total 5 0. 1.5 V 1.5 V R 5 17 V r total + + Energy 5 Pt . I 2 R . r total 1 9 . 1 3 R eq 5 R / 9. 1 R eq 5 3 1 1 R / 3 2 5 9 R . 1 R eq 5 1 R 1 1 1 R 2 1 1 R 3 . R / 3. R 5 r L A P 5 1 120 V 21 1.23 A 2 5 148 W P 5 VI 5 1 120 V 21 1.35 A 2 5 162 W T 5 20°C 1 217 C° 5 237°C T 2 T 0 5 I 0 2 I T a I T 5 1.35 A 2 1.23 A 1 1.23 A 21 4.5 3 10 2 4 1 2 2 1 2 5 217 C°. I 0 5 I T 1 1 1 a 3 T 2 T 0 42 . V I T 5 V I 0 1 1 1 a 3 T 2 T 0 42 . P 5 VI . R 5 V I . R T 5 R 0 1 1 1 a 3 T 2 T 0 42 .