cahier physique - physics workbook

PART I:ELECTRIC POWER, ELECTRIC ENERGY

1) the amount of energy used by an electric device depends on the resistance of the device. (motor, lamp, bell, TV, stereo ..).
P= V I V is the potential difference = energy per unit of charge (1C)
I is the current = number of unit of charges (number of 1C) per second
P is the power = the energy delivered to the device in 1 second (rate of change of energy)l
The units for the power P is _________, the units for the voltage V is ______, the units for the current I is __________. You know V = ____ (Ohm's law) so P = _________. (use only R and I).

2) A 15 Ω electric heater operates on a 120V outlet.
A) What is the current through the heater ?
(use Ohm's law, V = RI)
B) What is the power delivered by the heater
(used P = RI²)
C) How much energy is used by the heater in 30 seconds?
Energy = Power x time because power = energy delivered by second.

3) A 30 Ω resistor is connected to a 60V battery.
A) What is the current in the circuit ?

B) How much energy is used by the resistor in 5 minutes?
(hint: convert to seconds)

4) A 100W light bulb is 20% efficient. THat means that 20% of the electric energy is converted to light energy.
A) How many joules the light convert into light each minute of its operation?
(hint: find the energy delivered to the bulb during 1 minute, then find the useful energy (20% of it)

B) How many joules of heat does the light bulb produce each minute ?
(of course, that will be the remaining 80%, that's why you want to buy fluorescent bulb vs incandescent lamp, TO SAVE MONEY and our resources !)

5) The resistance of an electric stove element at operating temperature is 11 Ω.
A) 220V are applied across it. What is the current through the element? (Ohm's law)

B) How much energy does the element use in 30s ? (by now, you should know, P = RI² , and E = P t)

C) The element is being used to heat a kelle containing 1.20kg of water.
Assume that 70% of the heat is absorbed by the water. FInd the increase in temperature DT
You need to use the formula:
energy absorbed by water = heat absorbed = Q = s m DT
s =4.15J/g/C is the specific heat of water, m is the mass of the water being heated (in grams),
DT is the change in temperature (C)
it means it takes 4.15J to raise the temperature of a 1g of material by 1 degree
Q is the energy absorbed by the water that is 70% of the energy computed in B)

6) An electric heater is rated at 500W.
A) How much energy is delivered to the heater in half an hour?

B) The heater is being used to heat a room containing 50kg of air. The specific heat of air is s=1.10kJ/kg/C
(that is it takes 1100 J to increase the temperature of 1 kg of air by 1 degree). If 50% of the thermal energy heats the air in the room, what is the change in air temperature? (use Q = smDT and Q is 50% of the energy found in A) )
(keep kilograms, convert KJ to J)

7)A) How much energy does a 60W light bulb use in half an hour ?

B) If the light bulb is 12% efficient, how much heat it generates during the half hour?
(12% efficient means you waste 88% of energy. 88% of the energy delivered = heat = the bulb gets hot)

PART II: TRANSMISSION OF ELECTRIC ENERGY, the kilowatt hour

1) electricity must be carried from a power plant to your house using wires. All wires have resistance. They get hot and some the electric energy is wasted in heat. The energy wasted (for the consumers and for the power plant) depends on the current and on the resistance of the wires. P = RI²A) engineers want to keep this so called "I²R"loss as low as possible. REducing the I²R loss can be done in 2 ways:
i) _______________________________________________________________________________

ii) ______________________________________________________________________________
B) Of course you want to carry the same power P = VI (amount of energy per second) to the consumer.
So if you decrease I, you need to increase __________.
That's why we use very high voltage transmission lines.

C) When trying to reduce the I²R loss , why is it more important to reduce I than to reduce R ?

2) Electric companies have their own unit for energy. They measure the energy delivered to a consumer in kilowatt hour. The energy represented by 1000 watts delivered continuously for 3600s , which is one hour. It is, therefore:
1kWh = ______________ J

ENERGY in KWh = POWER in KW x TIME in hours
1KW = 1,000 W

3) A new color TV set draws 2A when operated on 120V.
A) How much power does the set use in Watts ? in KW ? (divide by 1,000)

B) If the set is operated for an average of 7 hours per day, what energy in KWh does it consume per month (30 days) ? (energy in kwh = power in kw x time in hours)

C) at $0.15 per kWh, what is the cost of operating the set per month ?

4) An electric space heater draws 15A from a 120V source. It is operated, on the average , for 5 h each day.
A) How much power does the heater use ?

B) How much energy in kWh does it consume in 30 days ?

C) at $0.15 per kWh, what does it cost the operate the heater in 30 days ?

5) A digital clock has an operating resistance of 12 000 Ω and is plugged into a 115V outlet.
A) How much current does it draw ?

B) How much power does it use?

C) If the owner of the clock pays $0.20 per kWh, what does it cost to operate the clock for 30 days ?

REVIEW

1) The damage caused by electric shock depends on the current flowing through the body. 1 mA (10^-3 A) can be felt. 5 mA (5 10^-3A) is painful. Above 15 mA, a person loses muscle control; 70 mA can be fatal.
A person with dry skin has a resistance from one arm to the other of about 10⁵ Ω . When skin is wet, the resistance drops to about 5 10³ V .
A) what is the minimum voltage placed across the arms that would produce a current that could be felt by a person with dry skin?

B) What effect would the same voltage by a person with wet skin ?

C) what would be the minimum voltage that would produce a current that could be felt when the skin is wet ?

2) A transistor radio operates by mean of a 9V battery that supplies it with 50 milliamperes (0.05A) of current.
A) If the cost of the battery is $1.20 and it lasts for 300 hours , what is the cost per kWh to operate the radio in this matter
(hint: first find the energy in Kwh = power in kw x time in hours. Then find the price.)

B) The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays $0.20 per kWh. What does it now cost to operate the radio for 300 hours ?

3) An electric motor operates a pump that irrigates a farmer's crop by pumping 10 000 L (1L = 1kg) of water a vertical distance of 8m into a field each hour (seconds). THe motor has an operating resistance of 22Ω and is connected across a 110V source.
A) What current does it draw ?
(Ohm's law)

B) How efficient is the motor ?
efficiency = [energy used to lift the water (useful energy) / Energy delivered by the voltage (electric energy) / ] x100
or efficiency = (energy out / energy in) x 100 =
hint: energy out = potential energy used to lift a mass of 10,000 kg , 8 meter. energy in = electric energy = power x time

4) Draw a schematic diagram to show a circuit that includes a 90V battery, an ammeter, and a resistance of 45 Ω connected in series. What is the ammeter reading ? Draw arrows showing the direction of conventional ccurrent.
Add a voltmeter to read the voltage across the resistance. What is its reading?