Problemas resueltos capitulo 28 paul e. tippens

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Chapter 28. Direct-Current Circuits
Resistors in Series and Parallel (Ignore internal resistances for batteries in this section.)
28-1. A 5-Ω resistor is connected in series with a 3-Ω resistor and a 16-V battery. What is the effective resistance and what is the current in the circuit? Re = R1 + R2 = 3 Ω +5 Ω; Re = 8.00 Ω [pic] I = 2.00 A

28-2. A 15-Ω resistor is connected in parallel with a 30-Ω resistor and a 30-V source of emf. What is the effective resistance and what total current is delivered? [pic]; Re = 10.0 Ω [pic]; I = 3.00 A

28-3. In Problem 28-2, what is the current in 15 and 30-Ω resistors?
For Parallel: V15 = V30 = 30 V;
…ver más…
28-20, what is the current and voltage across the 2-Ω resistor? Re = 2.22 Ω; [pic] Note that V5 = 12 V; [pic] V3,6 = (2.4 A)(2 Ω) = 4.80 V; [pic] I2 = I1 = 1.60 A; V2 = (1.6 A)(2 Ω) = 3.20 V I2 = 1.60 A; V2 = 3.20 V

EMF and Terminal Potential Difference
28-16. A load resistance of 8 Ω is connected in series with a 18-V battery whose internal resistance is 1.0 Ω. What current is delivered and what is the terminal voltage?
[pic]; I = 2.00 A

28-17. A resistance of 6 Ω is placed across a 12-V battery whose internal resistance is 0.3 Ω. What is the current delivered to the circuit? What is the terminal potential difference?
[pic] I = 1.90 A
VT = E – Ir = 12 V – (1.90 A)(0.3 Ω); VT = 11.4 V

28-18. Two resistors of 7 and 14 Ω are connected in parallel with a 16-V battery whose internal resistance is 0.25 Ω. What is the terminal potential difference and the current in delivered to the circuit? [pic]; Re = 0.25 Ω + 4.67 Ω [pic]; I = 3.25 A VT = E – Ir = 16 V – (3.25 A)(0.25 Ω);
VT = 15.2 V; I = 3.25 A

28-19. The open-circuit potential difference of a battery is 6 V. The current delivered to a 4-Ω resistor is 1.40 A. What is the internal resistance?
E = IRL + Ir; Ir = E - IRL
[pic]; r = 0.286 Ω

28-20. A dc motor draws 20 A from a 120-V dc line. If the internal resistance is 0.2 Ω, what is

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