# 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 A28-2. 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; [pic] I15 = 2.00 A
[pic] I30 = 1.00 ANote: I15 + I30 = IT = 3.00 A

28-4. What is the equivalent resistance of 2, 4, and 6-Ω resistors connected in parallel?

Re = 2 Ω + 4 Ω + 6 Ω; Re = 12.0 Ω
28-5. An 18-Ω resistor and a 9-Ω resistor are first connected in parallel and then in series with a 24-V battery. What is the effective resistance for each connection? Neglecting internal resistance, what is the total currentdelivered by the battery in each case?
[pic]; Re = 6.00 Ω
[pic]; I = 4.00 A
Re = R1 + R2 = 18 Ω +9 Ω; Re = 27.0 Ω
[pic] I = 0.889 A

28-6. A 12-Ω resistor and an 8-Ω resistor are first connected in parallel and then in series with a 28-V source of emf. What is the effectiveresistance and total current in each case?
[pic]; Re = 4.80 Ω
[pic]; I = 5.83 A
Re = R1 + R2 = 12 Ω +8 Ω; Re = 20.0 Ω [pic] I = 1.40 A

28-7. An 8-Ω resistor and a 3-Ω resistor are first connected in parallel and then in series with a 12-V source. Find the effective resistance and total current for each connection?[pic]; Re = 2.18 Ω [pic]; I = 5.50 A
Re = R1 + R2 = 3 Ω +8 Ω; Re = 11.0 Ω [pic] I = 1.09 A

28-8. Given three resistors of 80, 60, and 40 Ω, find their effective resistance when connected in series and when connected in parallel.
Series: Re = 80 Ω + 60 Ω + 40 Ω ; Re = 180 Ω
Parallel: [pic] Re = 18.5 Ω

28-9. Three resistancesof 4, 9, and 11 Ω are connected first in series and then in parallel. Find the effective resistance for each connection.
Series: Re = 4 Ω + 9 Ω + 11 Ω ; Re = 24.0 Ω
Parallel: [pic] Re = 2.21 Ω

*28-10. A 9-Ω resistor is connected in series with two parallel resistors of 6 and 12 Ω. What is the terminal potential difference if the total current from the battery is 4 A?[pic]
VT = IR = (4 A)(13 Ω); VT = 52.0 V

*28-11. For the circuit described in Problem 28-10, what is the voltage across the 9-Ω resistor and what is the current through the 6-Ω resistor?
V9 = (4 A)(9 Ω) = 36 V; V9 = 36.0 V
The rest of the 52 V drops across each of the parallel resistors:
V6 = V7 = 52 V – 36 V; V6 = 16 V
[pic] I6 = 2.67 A*28-12. Find the equivalent resistance of the circuit drawn in Fig. 28-19.
Start at far right and reduce circuit in steps: R’ = 1 Ω + 3 Ω + 2 Ω = 6 Ω;
[pic]; Re = 2 Ω + 4 Ω + 2 Ω ; Re = 8 Ω

*28-13. Find the equivalent resistance of the circuit shown in Fig. 28-20.
Start at far right and reduce circuit in steps: R = 1 Ω + 2 Ω = 3 Ω;
[pic]; R’’ = 2 Ω + 3 Ω = 5 Ω
[pic]; Re =2.22 Ω

*28-14. If a potential difference of 24 V is applied to the circuit drawn in Fig. 28-19, what is the current and voltage across the 1-Ω resistor?
Re = 8.00 Ω; [pic]
The voltage across the 3 and 6-Ω parallel
connection is found from It and the 2-Ω combination resistance:
V3 = V6 = (2 Ω)(3.00 A); V6 = 6.00 V; [pic]
Thus, I1 = I6 = 1.00...