Download Chapter 7: Current Electricity End of Chapter Questions

Chapter 7: Current Electricity
End of Chapter Questions
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What condition is necessary for the sustained flow of water in a pipe? What analogous condition is necessary for
the sustained flow of charge in a wire?
Why are electrons, rather than protons, the principal charge carriers in metal wires?
What exactly is an ampere?
Why is a current-carrying wire normally not electrically charged?
Name two kinds of "electric pumps."
How much energy is supplied to each coulomb of charge that flows through a 12-V battery?
Does charge flow through a circuit or into a circuit? Does voltage flow through a circuit, or is voltage established
across a circuit?
Will water flow more easily through a wide pipe or a narrow pipe? Will current flow more easily through a thick
wire or a thin wire?
How does wetness affect the resistance of your body?
For a given voltage, what happens to the amount of current that flows in your skin when you perspire?
Why is it a very poor idea to handle electrical devices while in the bathtub?
Distinguish between dc and ac.
Does a battery produce dc or ac? Does the generator at a power station produce dc or ac?
What is the relationship among electric power, current, and voltage?
Which of these is a unit of power and which is a unit of energy: a watt; a kilowatt; a kilowatt-hour?
Distinguish between a kilowatt and a kilowatt-hour.
What is an electric circuit?
In a circuit of two lamps in series, if the current through one lamp is 1 A, what is the current through the other
lamp? Defend your answer.
What is a main shortcoming of a series circuit?
In a circuit of two lamps in parallel, if there are 6 V across one lamp, what is the voltage across the other lamp?
How does the sum of the currents though the branches of a simple parallel circuit compare to the current that
flows through the voltage source?
As more lines are added at a fast food restaurant, the resistance to people getting served is reduced. How is this
similar to what happens when more branches are added to a parallel circuit?
Are household circuits normally wired in series or in parallel, and when are they overloaded?
What is the function of fuses or circuit breakers in a circuit?
The wattage marked on a light bulb is not an inherent property of the bulb but depends on the voltage to which it
is connected, usually 110 or 120 V. How many amperes flow through a 60-W bulb connected in a 120-V circuit?
Rearrange the equation: Current = Voltage / Resistance, to express resistance in terms of current and voltage.
Then solve the following: A certain device in a 120-V circuit has a current rating of 20 A. What is the resistance
of the device (how many ohms)?
Using the equation: Power = Current x Voltage, find the current drawn by a 1200-W hair dryer connected to 120
V. Then using the method you used in the previous problem, find the resistance of the hair dryer.
The total charge that an automobile battery can supply without being recharged is given in terms of amperehours. A typical 12-V battery has a rating of 60 ampere-hours (60 A for 1 hour, 30 A for 2 hours, and so on).
Suppose you forget to turn off the headlights in your parked automobile. If each of the two headlights draws 3 A,
how long will it be before your battery is “dead”?
How much does it cost to operate a 100-W lamp continuously for 1 week if the power utility rate is 15 cents/kWh?
A 4-W night light is plugged into a 120-V circuit and operates continuously for 1 year. Find the following: (a) the
current it draws, (b) the resistance of its filament, (c) the energy consumed in a year, and (d) the cost of its
operation for a year at the rate of 15 cents/kWh.
An electric iron connected to a 110-V source draws 9 A of current. How much heat in joules does it generate in a
minute?
How many coulombs of charge flow through the iron in the previous problem in one minute?
A certain light bulb with a resistance of 95 ohms is labeled “150 W”. Was this bulb designed for use in a 120-V
circuit or a 220-V circuit?
In periods of peak demand, power companies lower their voltage. This saves them power (and saves you
money!). To see the effect, consider a 1200-W toaster that draws 10 A when connected to 120 V. Suppose the
voltage is lowered by 10% to 108 V. By how much does the current decrease? By how much does the power
decrease? (Caution: The 1200-W label is valid only when 120 V is applied. When the voltage is lowered, it is
the resistance of the toaster, not its power, that remains constant.)
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Sustained water flow requires sustained pressure difference by a pump. Analogously, a sustained potential difference must
be maintained for a sustained flow of charge.
Because electrons in the metal are free to move throughout the atomic lattice. Protons, however, are locked in the atomic
lattice.
A unit of current. One ampere is a rate of flow equal to 1 coulomb of charge per second.
Because there are as many electrons as protons in the wire.
Generators or chemical batteries.
12 joules.
Charge flows through a circuit. Voltage is established across a circuit.
Water flows with less resistance through wide pipes. Similarly, charge flows with less resistance through thick wires.
Wetness reduces body resistance.
Current increases because resistance decreases.
Your overall resistance is lowered, which means greater chance of harmful current in your body if a voltage is established
across it.
Direct current, dc, flows in one direction. Alternating current, ac, alternates back and forth in direction.
A battery produces dc. A generator produces ac.
Power = current x voltage.
A kilowatt-hour is a unit of energy (energy = power x time).
A kilowatt is a unit of power, and a kilowatt-hour is a unit of energy. Again, energy = power x time.
Any path along which electrons can flow.
Also 1 A, as current in all elements in series is the same.
If one device fails, current in the whole circuit ceases.
Also 6 V, as voltage across parallel branches is the same.
The sum of the currents in parallel branches is equal to the total current in the circuit, which is what flows through the
voltage source.
As more branches are added, the overall resistance is likewise reduced.
Parallel. They are overloaded when a buildup of current overheats them.
To break the circuit when unsafe amounts of current build up.
25. From "Power = current x voltage," 60 watts = current x 120 volts, current =
26. From current =
, resistance =
= 6 Ohms.
27. From power = current x voltage, current =
= 10 A.
= 0.5 A.
From the formula derived above, resistance =
= 12 Ohms.
28. Two headlights draw 6 amps, so the 60 ampere-hour battery will last for about 10 hours.
29. $2.52. First, 100 watts = 0.1 kilowatt. Second, there are 168 hours in one week (7 days x 24 hours/day = 168 hours). So
168 hours x 0.1 kilowatt = 16.8 kilowatt-hours, which at 15 cents per kWh comes to $2.52.
30. (a) From power = current x voltage, current = power/voltage = 4W/120V = 1/30 A.
(b) From current = voltage/resistance (Ohm's law), resistance = voltage/current = 120 V/(1/30 A) = 3600 Ohms.
(c) First, 4 watts = 0.004 kilowatt. Second, there are 8760 hours in a year (24 hours/day x 365 days = 8760 hours). So
8760 hours x 0.004 kilowatt = 35.0 kWh.
(d) At the rate of 15 cents per kWh, the annual cost is 35.0 kWh x $0.15/kWh = $5.25.
31. The iron's power is P = IV = (110 V)(9 A) = 990 W = 990 J/s. The heat energy generated in 1 minute is E = power x time =
(990 J/s)(60 s) = 59,400 J.
32. Since current is charge per unit time, charge is current x time: q = It = (9 A)(60 s) = (9 C/s)(60 s) = 540 C. (Charges of this
magnitude on the move are commonplace, but this quantity of charge accumulated in one place would be incredibly large.)
33. It was designed for use in a 120-V circuit. With an applied voltage of 120 V, the current in the bulb is I = V/R = (120 V)/(95
W) = 1.26 A. The power dissipated by the bulb is then P = IV = (1.26 A)(120 V) = 151 W, close to the rated value. If this
bulb is connected to 220 V, it would carry twice as much current and would dissipate four times as much power (twice the
current x twice the voltage), more than 600 W. It would likely burn out. (This problem can also be solved by first carrying
out some algebraic manipulation. Since current = voltage/resistance, we can write the formula for power as P = IV = (V/R)V
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= V /R. Solving for V gives V = vPR. Substituting for the power and the resistance gives V = v(150)(95) = 119 V.)
34. The resistance of the toaster is R = V/I = (120 V)/(10 A) = 12 ohms. So when 108 V is applied, the current is I = V/R = (108
V)/(12 W) = 9.0 A and the power is P = IV = (9.0 A)(108V) = 972 W, only 81 percent of the normal power. (Can you see the
reason for 81 percent? Current and voltage are both decreased by 10 percent, and 0.9 x 0.9 = 0.81.)