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electricity review answers (1P)

4. The charge on an object can be detected by observing the interaction between it and an object of known charge. An electroscope is typically used for this purpose. First, the electroscope is charged with an object of known charge, and then the object with unknown charge is brought near the electroscope. If the leaves of the electroscope separate from one another further, then the unknown charge is the same as that which was used to charge the electroscope. If the leaves of the electroscope approach one another, then the unknown object has the opposite charge to the charge of the known object.

5. Students’ diagrams should indicate a charged object inducing charge separation in a neutral object, causing an attractive force between the charged object and the locally charged area of the neutral object.

7. (a) Balloon B is also negatively charged.

(b) If the balloons attract one another, then balloon B could be positive or neutral, since negatively charged balloon A will attract both neutral and positively charged objects.

8. (a) Object D has a positive charge.

(b) Object C has a negative charge.

10. Students’ diagrams should be well labelled and indicate the process of lightning forming, specifically showing charge separation in the clouds, the induction of a locally positive area on Earth, the formation of a channel made of charged particles, and the transfer of excess electrons from the cloud to the ground. See Figure 10.22 on student book page 410 for an example.

11. (a) Clothing from the clothes dryer often sticks together because of having been rubbed against other clothing made of different materials. This friction causes electrons to be transferred and charges to be accumulated.

(b) This effect can be reduced by removing clothing from the dryer before it is fully dry, drying only articles made of the same material together, and/or using anti-static sheets in the dryer.

17.(a) The four factors affecting resistance in a wire are material, temperature, length, and cross-sectional area.

(b) Both increased length and temperature increase the resistance of a wire. Increased cross-sectional area decreases resistance. The material the wire is made from also affects the resistance of the wire depending on how good a conductor the material is.

20.

(c) More current flows in the parallel circuit.

(d) Both bulbs will go dark if one of the bulbs in circuit (a) is removed.

(e) There will be no difference in brightness in the remaining bulb in circuit (b) if one of the bulbs is removed.

21. (a) Voltage V1 in the circuit is 6.0 V.

(b) Current A1 in the circuit is 2.0 A.

(c) It is a series circuit.

22. (a) Voltage V1 in the circuit is 3.0 V.

(b) Current A1 in the circuit is 1.0 A.

(c) It is a parallel circuit.

26. Homes are wired for safety with circuit breakers or fuses, grounded plugs, and ground fault circuit interrupters.

27. Fused safety power bars help prevent damage to sensitive electronics due to power surges or voltage spikes from lightning strikes or the electrical grid.

35. Electrical energy consumption is typically measured in kilowatthours (kW•h).

39. The EnerGuide and Energy Star labels could be used to help determine which appliances are the most efficient and will result in the lowest long-term energy cost of operation.

48. The bird is not grounded and since it provides no path to ground, electric charge does not flow through it and it does not experience an electric shock.

51. (a) A large aluminum ladder may provide a path for current to flow to ground from the overhead power wire through the worker carrying the ladder.

(b) Removing the ground plug from the cord of an appliance breaks the safe pathway to ground for the appliance, potentially exposing the user to unsafe electrical discharges in case of a short circuit in the device.

(c) A frayed electrical cord on a washing machine is hazardous because it may create a short circuit, potentially leading to electric shock or fire.

(d) A piece of aluminum instead of a fuse will not provide overcurrent protection to the circuit, may melt, or possibly cause a fire. No other object should ever be substituted for a fuse.

52. Replacing a cord with one that is much thinner than the original is hazardous because it may overheat. This occurs due to the higher resistance of the smaller diameter cord than the original larger diameter one.

55. A more efficient appliance that costs more than a regular model makes economic sense if the electricity cost savings of the efficient appliance can recover the initial increased purchase price within a reasonable period of time and continue producing meaningful energy savings after that.

57.

Given

Voltage V = 6.0 V

Current I = 0.0020 A (2.0 mA is 0.0020 A, since there are 1000 A in 1 A)

Required

Resistance R = ?

Analysis and Solution

The correct equation is V = IR, rearranged to R =

V/I

Substitute the values and their units, and then solve the problem.

R = V/I

= 6.0 V / 0.0020 A

 = 3000 Ω

Paraphrase

The resistance of the resistor is 3000 Ω.

58. (a)

Given

Current I = 1.5 A

Resistance R = 5.0 Ω

Required

Voltage V = ?

Analysis and Solution

The correct equation is V = IR.

Substitute the values and their units, and then solve the problem.

V = IR

= (1.5 A)(5.0 Ω)

= 7.5 V

Paraphrase

The voltage across the resistor is 7.5 V.

(b)

Given

Voltage V = 80 V

Resistance R = 20 Ω

Required

Current I = ?

Analysis and Solution

The correct equation is V = IR, rearranged to I = V/R

Substitute the values and their units, and then solve the problem.

I = V/R

= 80V/20 Ω

= 4 A

Paraphrase

The current through the resistor is 4 A.

(c)

Given

Voltage V = 12 V

Current I = 240 A

Required

Resistance R = ?

Analysis and Solution

The correct equation is V = IR, rearranged to R = V/I

Substitute the values and their units, and then solve the problem.

R = V/I

= 12V/240 A

= 0.05 Ω

Paraphrase

The resistance of the motor is 0.05 Ω.

59.This question contains three 40-Ω resistors in series for a total series resistance of 120 Ω. Values in the table are found from the  equation I = V/R, using the voltage V from the first column of the table and the resistance R = 120 Ω.

Voltage (V) Current (A)

2.0 0.017

4.0 0.033

6.0 0.050

8.0 0.067

10.0 0.083

Device Input            Energy (kJ)    Output Energy (kJ)    Percent Efficiency (%)

Gas-powered SUV           675                     81                                12

Gas-electric hybrid car    675                   195                                29

Natural gas furnace         110 000             85 000                             77

Electric baseboard

heater                            9.5                    6.0                                63

Alkaline dry cell             84.52                 74.38                               88

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