Chapter 20 Pretest Circuits

1. If the batteries in a portable CD player provide a terminal voltage of 12 V, what is the potential difference across the entire player? A. 3.0 VB. 4.0 V C. 6.0 VD. 12 V

2. Three resistors with values of 3.0 Ω, 6.0 Ω, and 12 Ω are connected in series. What is the equivalent resistance of this combination? A ΩB. 1.7 Ω C. 7.0 ΩD. 21 Ω

3. Three resistors connected in parallel carry currents labeled I 1, I 2, and I 3. Which of the following expresses the total current I T in the combined system? A. I T = I 1 + I 2 + I 3 B. I T = (1/I 1 + 1/I 2 + 1/I 3 ) C. I T = I 1 = I 2 = I 3 D. I T = (1/I 1 + 1/I 2 + 1/I 3 ) -1

4. What is the equivalent resistance for the resistors in the figure above? A. 25 ΩB. 10 Ω C. 7.5 Ω D. 5.0 Ω

WOW, no figure! Sorry!

1. What are the two formulas for calculating equivalent resistance?

Series: R T = R 1 + R 2 + R 3 Parallel: 1/R T = 1/R 1 + 1/R 2 + 1/R 3

2a. What is the equivalent resistance in the circuit shown above? b. If the battery is 12V, what is the current in the circuit? c. What is the current in the 6Ω resistor? d. What is the rate of energy dissipation for the 6Ω resistor? e. What is the rate of energy dissipation for the entire circuit?

2a. The 8 Ω and 4 Ω resistors are replaced with a 12 Ω resistor. This 12 and the other 12 are in parallel, so they can be replaced with a 6 Ω resistor. The 6 Ω and 4 Ω resistors are in parallel, so they can be replaced with a 2.4 Ω resistor. This 2.4 Ω resistor is in series with the 6 Ω resistor, so the total resistance is 8.4 Ω.

b. If the battery is 12V, what is the current in the circuit? V = IR 12 V = I x 8.4 Ω I = 1.43 A

c. What is the current in the 6Ω resistor? The 6 Ω resistor is in series with the source, so it gets all the current. I = 1.43 A

d. What is the rate of energy dissipation for the 6Ω resistor? The current I is 1.43 A. The voltage drop is V = IR, V = 1.43 x 6 = 8.58 V. P = IV P = 1.43 x 8.58 P = 12.3 W

e. What is the rate of energy dissipation for the entire circuit? P = IV P = 1.43 x 12 P = W

3. A student is provided with a 12.0-V battery of negligible internal resistance and four resistors with the following resistances: 100 Ω, 30 Ω, 20 Ω, and 10 Ω. The student also has plenty of wire of negligible resistance available to make the connections as desired. (a)Using all of these components, draw a circuit diagram in which each resistor has nonzero current flowing through it, but in which the current from the battery is as small as possible.

A series circuit would make the resistance highest, so the current would be lowest.

3. A student is provided with a 12.0-V battery of negligible internal resistance and four resistors with the following resistances: 100 Ω, 30 Ω, 20 Ω, and 10 Ω. The student also has plenty of wire of negligible resistance available to make the connections as desired. (b) Using all of these components, draw a circuit diagram in which each resistor has nonzero current flowing through it, but in which the current from the battery is as large as possible (without short circuiting the battery).

A parallel circuit would make the resistance lowest, so the current would be as high as possible.

The battery and resistors are now connected in the circuit shown above. (c) Determine the equivalent resistance in the circuit.

(c) Determine the equivalent resistance in the circuit. The 20 Ω and 30 Ω resistors are in series, so they are equal to a 50 Ω resistor. This 50 is in parallel with the 100 Ω resistor, so they can be replaced with a 33 Ω resistor. This 33 is in series with the 10, so the total resistance is 43 Ω.

(d) Determine the current in the 10-Ω resistor. This resistor is in series with the source, so it gets all the current. V = IR 12 V = I x 43 Ω. I = 0.28 A

4. Three resistors with values of 16 Ω, 19 Ω, and 25 Ω, respectively, are connected in parallel. What is their equivalent resistance?

1/R T = 1/R 1 + 1/R 2 + 1/R 3 1/R T = 1/16 + 1/19 + 1/25 1/R T = /R T = R T = 6.45 Ω

1/R P = 1/R 1 + 1/R 2 + 1/R 3 1/R P = 1/3 + 1/4 + 1/5 1/R P = /R P = 0.78 R P = 1.28 Ω