We start with the basic definition of power, P = W / t, then, we introduce the different expressions that we have already developed.
Using the result, W = qV and q = It ⇔ t = q / I , we have P qV q I --- ⎝ ⎠ ⎛ ⎞ --- IV = =
The same result could have been obtained using the fact that the electrical energy,
W = q V = ItV so that P W t --- ItV t --- IV = = =
However, we also have that V = IR , so that
P = I×IR = I2R
We also could have used I = V / R , giving the result,
P IV V R ---×V V 2 R --- = = = Summary P W t --- I2R V VI 2 R --- = = = =
The commercial unit of electrical energy is the kilowatt-
hour (kW h). It is the energy consumed when 1 kW of
power is used for one hour. The consumer has to pay a certain cost per kilowatt-hour say 14 cents per kW h.
The heating effect of a current was investigated in 1841 by James Joule. He was able to demonstrate that by supplying electrical energy to a high resistance coil of wire this energy could be converted to thermal energy.
V×I×t = m×c×ΔT
Example
An electrical appliance is rated as 2.5 kW, 240 V.
(a) Determine the current needed for it to operate.
(b) Calculate the energy it would consume in 2.0 hours.
Solution
(a) Given that P = 2.5 × 103 and V = 240 V, we
use the formula, P = IV. Now,
the current drawn is 1.0 × 101 A.
(b) Next, we use the formula W = VIt, so that
W = (240 V)×(10.4 A)×(7.2×103 s)
1.8×107 J
=
The energy consumed is 1.8 × 107J.
Example
A 2.5 kW blow heater is used for eight hours. Calculate the cost if electricity is sold at 12 cents per kilowatt-hour?
P IV⇒I P V --- 2.5 10 3 × 240 --- 10.4 = = = =
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Solution
Using the fact that energy consumed (E) = power × time, we have
E = (2.5 kW)×(8 h) = 20 kW h
Therefore, the Cost = (20 kW h) × $0.12 per kW h = $2.40 The cost to run the heater is two dollars forty cents ($2.40).
Example
A 1.2 kW electric water heater that is made of aluminium is used to heat 2.5 L of water from 25 °C to the boil. If the mass of aluminium water heater is 350 g and all the electrical energy is converted to heat energy, determine the time in minutes to bring the water to the boil.
(Specific heat capacity of aluminium is 9.1 × 102 J kg-1K-1)
Solution
A 1.2 kW heater delivers 1200 joules each second. 2.5 L = 2.5 kg
The electrical energy is used to heat the heater and the water from 25 °C to 100 °C.
Q = mcΔTheater + mcΔTwater
= (0.35 kg × 9.1 × 102 J kg-1K-1 × 75 °C) +
(2.5kg × 4180 J kg-1K-1 × 75 °C)
= 8.08 × 105 J
Now if the heater delivers 1200 J per second, then to consume 8.08 × 105 J will take: 8.08 × 105 J _________ 1200 J = _____ 673 s 60 = 11.2 min ≈ 11 min Exercise 5.1
1. When a current is flowing through a wire attached to a dry cell
A. positive charges flow from negative to positive terminal
B. positive charges flow from positive to negative terminal
C. negative charges flow from negative to positive terminal
D. negative charges flow from positive to negative terminal
2. The total charge passing the same point in a conductor during 2.0 μs of an electric current of 7.5 mA is
A. 15 nC B. 15 mC C. 15 μC D. 1.5 × 104 μC
3. An electric heater raises the temperature of a measured quantity of water. 6.00 × 103 J of energy
is absorbed by the water in 30.0 seconds. What is the minimum power rating of the heater?
A. 5.00 × 102 W
B. 2.00 × 102 W
C. 2.00 × 103 W
D. 1.80 × 105 W
4. If the power developed in an electric circuit is doubled, the energy used in one second is
A. halved B. doubled C. quartered D. quadrupled
5. The electron volt is defined as
A. the energy acquired by an electron when it passes through a potential difference of 1.0 V. B. the voltage of an electron.
C. a fraction of the ionisation of an electron. D. unit of energy exactly equal to 1.6 × 10–19 J.
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6. The definition of the unit of current, the ampere, is based on
A. the charge per unit time delivered by an emf of 1.0 V.
B. the force per unit length on parallel current-carrying wires.
C. the force per unit length on a conductor in a magnetic field
D. the charge passing a point per unit time.
7. Identify the charge-carriers in
a. a length of copper wire.
b. an aqueous solution of sodium chloride. c. the atmosphere during a lightning storm.
8. The speed with which electrons move through a copper wire is typically 10-4 m s-1.
a. Explain why is it that the electrons cannot travel faster in the conductor?
b. Explain why the electron drift produces heat?
9. Calculate the resistance of a wire if 0.5 V across it causes a current of 2.5 A to flow.
10. Calculate current flow through a 20 MΩ resistor connected across a 100 kV power supply.
11. A thin copper wire 200 cm in length has a 9 V dry cell connected between its ends. Determine the voltage drop that occurs along 30 cm of this wire.
12. Determine the length of tungsten wire with a diameter of 1.0 mm that is used to make a 20.0 Ω resistor.
13. A nichrome wire has a diameter of 0.40 mm. Calculate the length of this wire needed to carry a current of 30 mA when there is a potential difference of 12 V across it.
14. Explain in terms of atomic and electron movement, why resistance increases with temperature.
15. Determine how many coulombs there are when 2.0 A flows for 2.0 hours?
16. Distinguish the difference between an ohmic and a non–ohmic material.
17. Copy out and complete the table
Appliance Power (Watt) p.d (Volt) Current (Ampere)
Fuse rating needed (3,5,10,13 A) Digital clock 4 240 Television 200 240 Hair dryer 110 5 Iron 230 4 Kettle 240 10
18. Calculate the cost of heating the water to wash the dishes if the sink is 48 cm long, 25 cm wide and the water is 25 cm high. The tap water is at 14 °C and the final temperature before washing up is 62 °C. Power is sold to the consumer at 14 cents per kilowatt-hour.
19. The element of an electric jug has a resistance of 60 Ω and draws a current of 3.0 A. Determine by how much the temperature of 5.0 kg of water will rise if it is on for 6 minutes.
20. Calculate the cost to heat 200 kg of water from 12°C to its boiling point if power costs 14 cents per kilowatt-hour.
21. Determine the work done in moving a charge of 10.0 nC through a potential difference of 1.50 × 102 V?
22. An electron in an electron gun of a picture tube is accelerated by a potential 2.5 × 103 V. Calculate the
kinetic energy gained by the electron in electron- volts.
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5.2.1 Define electromotive force (emf).
5.2.2 Describe the concept of internal resistance.
5.2.3 Apply the equations for resistors in series and in parallel.
5.2.4 Draw circuit diagrams.
5.2.5 Describe the use of ideal ammeters and ideal voltmeters.
5.2.6 Describe a potential divider.
5.2.7 Explain the use of sensors in potential divider circuits.
5.2.8 Solve problems involving electric circuits.
© IBO 2007
5.2.1 ELECTROMOTIVE
FORCE
In the previous section, potential difference was defined in terms of the amount of work that has to be done per unit charge to move a positively-charged body in an electric field. V Δ ΔW q Δ --- or WΔ ( ) VΔq Δ = =
From the present definition of electric current, we have,
I Δq t
Δ
--- or qΔ I tΔ
= =
By substituting for q in the top equation, we end up with
W
Δ I×Δt×ΔV or VΔ ΔW
I tΔ
---
= =
Therefore it can be stated that
‘Potential difference in external circuits is the power, (P), dissipated (released) per unit current’
Either definition for potential difference is acceptable. However, the above-mentioned ties voltage and current together.
Note: It is common place to use the expression V = W / It as opposed to ΔV = ΔW / IΔt and q = It as opposed to q = IΔt. That is, to leave out the ‘change in’ (or ‘Δ’) notation when solving problems.
The terms emf, potential difference and voltage are commonly interchanged in talking about electricity. The term electromotive force should be avoided as it is not a force at all but rather a potential energy difference. Again, there are historical reasons for the use of the term in the first place.
In the true sense, electromotive force (emf) is the work
per unit charge made available by an electrical source.
For the simple circuit in Figure 507 , the emf of the dry cell is 1.5 V. However, when a voltmeter is used to test the potential difference across the resistor the reading on the voltmeter is found to be less than 1.5 V. This is due to the cell not being an ‘ideal dry cell’ because it has some internal resistance.
For the moment, let us say that emf is the energy supplied per unit charge, and potential difference is the energy
released (dissipated) per unit charge.
Example
A battery supplies 15.0 J of energy to 4.00 C of charge passing through it. Determine the emf of the battery
Solution
Using the formula, V = W / q, we have, V = 15.0J / 4.00 C = 3.75 V
The emf of the battery is 3.75 V.