A continuous and closed path of an electric current is called an electric circuit. It usually has a cell, a switch, connecting wires and an electrical device like a bulb.
2Define the unit of current.
The SI unit of electric current is ampere (A). One ampere is the flow of one coulomb of charge per second.
1 A = 1 C / 1 s
3Calculate the number of electrons constituting one coulomb of charge.
Charge on one electron, e = 1.6 × 10-19 C Number of electrons n = Q/e = 1/(1.6 × 10-19) n = 6.25 × 1018 electrons
Intext Questions (Page 174)
1Name a device that helps to maintain a potential difference across a conductor.
Answer: A cell or a battery
2What is meant by saying that the potential difference between two points is 1 V?
It means that 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
1 V = 1 J / 1 C
3How much energy is given to each coulomb of charge passing through a 6 V battery?
W = V × Q = 6 V × 1 C = 6 J
Intext Questions (Page 181)
1On what factors does the resistance of a conductor depend?
Length of the conductor (R is directly proportional to length).
Area of cross-section (R is inversely proportional to area).
Nature of the material of the conductor.
Temperature of the conductor.
2Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Current will flow more easily through a thick wire. Resistance is inversely proportional to the area of cross-section. A thick wire has a larger area, so it has less resistance.
3Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
By Ohm’s law, I = V/R. If V becomes half and R stays the same, the current also becomes half.
4Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Alloys have higher resistivity than pure metals. So they produce more heat. Alloys also do not oxidise (burn) easily at high temperatures. So they are used in the coils of toasters and irons.
5Use the data in Table 11.2 to answer the following – (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?
(a) Iron – its resistivity (10.0 × 10-8 Ω m) is less than that of mercury (94.0 × 10-8 Ω m).
(b) Silver – it has the lowest resistivity (1.60 × 10-8 Ω m).
Intext Questions (Page 185)
1Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Battery of three 2 V cells with 5 Ω, 8 Ω and 12 Ω resistors and a plug key in series.
2Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Ammeter in series with the resistors; voltmeter in parallel across the 12 Ω resistor.
Total resistance R = 5 + 8 + 12 = 25 Ω Voltage V = 3 × 2 = 6 V Ammeter reading I = V/R = 6/25 = 0.24 A Voltmeter reading V = I × R = 0.24 × 12 = 2.88 V
Intext Questions (Page 188)
1Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 106 Ω, (b) 1 Ω and 103 Ω, and 106 Ω.
In parallel, the equivalent resistance is less than the smallest resistance.
(a) 1/R = 1/1 + 1/106 = 1.000001, so R ≈ 1 Ω (slightly less than 1 Ω) (b) 1/R = 1/1 + 1/103 + 1/106 = 1.001001, so R ≈ 0.999 Ω (about 1 Ω)
2An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
1/R = 1/100 + 1/50 + 1/500 = (5 + 10 + 1)/500 = 16/500 R = 500/16 = 31.25 Ω I = V/R = 220/31.25 = 7.04 A
The electric iron has a resistance of 31.25 Ω and draws a current of 7.04 A.
3What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Each device gets the full voltage of the source.
Each device can be switched on or off separately.
If one device fails, the others keep working.
The total resistance is less, so each device gets the current it needs.
4How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
(a) 4 Ω
Connect 3 Ω and 6 Ω in parallel. Then connect this combination in series with 2 Ω.
Rp = (3 × 6)/(3 + 6) = 2 Ω; total = 2 + 2 = 4 Ω
(b) 1 Ω
Connect all three in parallel.
1/R = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 1, so R = 1 Ω
5What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
1Why does the cord of an electric heater not glow while the heating element does?
The cord is made of copper, which has very low resistance. So very little heat is produced in it. The heating element is made of an alloy with high resistance. So a lot of heat is produced in it and it glows.
2Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
H = V × Q = 50 × 96000 H = 4.8 × 106 J
3An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
H = I2Rt = 52 × 20 × 30 H = 15000 J = 1.5 × 104 J
Intext Questions (Page 192)
1What determines the rate at which energy is delivered by a current?
The rate is determined by the electric power of the device. P = V × I. So it depends on the potential difference and the current.
2An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
P = V × I = 220 × 5 = 1100 W Energy = P × t = 1100 W × 2 h = 2200 Wh = 2.2 kWh In joules: 1100 × 7200 s = 7.92 × 106 J
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Exercises
1A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is – (a) 1/25 (b) 1/5 (c) 5 (d) 25
Answer: (d) 25
Each part = R/5. Five in parallel: R′ = (R/5)/5 = R/25 R/R′ = 25
2Which of the following terms does not represent electrical power in a circuit? (a) I2R (b) IR2 (c) VI (d) V2/R
Answer: (b) IR2
3An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be – (a) 100 W (b) 75 W (c) 50 W (d) 25 W
Answer: (d) 25 W
R = V2/P = (220 × 220)/100 = 484 Ω P = V2/R = (110 × 110)/484 = 25 W
4Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be – (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
Answer: (c) 1:4
Series: Rs = 2R. Parallel: Rp = R/2 H = V2t/R, so Hs : Hp = Rp : Rs = R/2 : 2R = 1 : 4
5How is a voltmeter connected in the circuit to measure the potential difference between two points?
A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured.
6A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
d = 0.5 mm = 5 × 10-4 m A = πd2/4 = 3.14 × (5 × 10-4)2/4 = 1.96 × 10-7 m2 R = ρl/A, so l = RA/ρ l = (10 × 1.96 × 10-7)/(1.6 × 10-8) = 122.7 m
If the diameter is doubled, the area becomes 4 times. So the resistance becomes one-fourth.
New R = 10/4 = 2.5 Ω
The resistance decreases by 7.5 Ω (it becomes 1/4 of the original).
7The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below – I (amperes): 0.5, 1.0, 2.0, 3.0, 4.0; V (volts): 1.6, 3.4, 6.7, 10.2, 13.2. Plot a graph between V and I and calculate the resistance of that resistor.
V–I graph: a straight line through the origin.
The graph is a straight line. So V is proportional to I. The slope of the line gives the resistance.
Taking the points (1.0 A, 3.4 V) and (3.0 A, 10.2 V): R = (10.2 − 3.4)/(3.0 − 1.0) = 6.8/2.0 = 3.4 Ω
8When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
I = 2.5 mA = 2.5 × 10-3 A R = V/I = 12/(2.5 × 10-3) = 4800 Ω = 4.8 kΩ
9A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Total R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω I = V/R = 9/13.4 = 0.67 A
In series, the same current flows through every resistor. So 0.67 A flows through the 12 Ω resistor.
10How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Required resistance R = V/I = 220/5 = 44 Ω For n equal resistors in parallel, R = 176/n 44 = 176/n, so n = 4
11Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
(i) 9 Ω
Connect two resistors in parallel. Connect the third in series with this combination.
6 × 6/(6 + 6) = 3 Ω; 3 + 6 = 9 Ω
(ii) 4 Ω
Connect two resistors in series. Connect the third in parallel with this combination.
6 + 6 = 12 Ω; 12 × 6/(12 + 6) = 72/18 = 4 Ω
Connecting three 6 Ω resistors to get (i) 9 Ω and (ii) 4 Ω.
12Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Current through one lamp = P/V = 10/220 = 1/22 A Number of lamps = 5 ÷ (1/22) = 110 lamps
13A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Separately: I = 220/24 = 9.17 A In series: R = 48 Ω, I = 220/48 = 4.58 A In parallel: R = 12 Ω, I = 220/12 = 18.33 A
14Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
(i) 6 V battery, 1 Ω and 2 Ω in series
I = 6/(1 + 2) = 2 A P = I2R = 22 × 2 = 8 W
(ii) 4 V battery, 12 Ω and 2 Ω in parallel
Voltage across 2 Ω = 4 V P = V2/R = 42/2 = 8 W
The power used in the 2 Ω resistor is the same (8 W) in both cases.
15Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Total power = 100 + 60 = 160 W I = P/V = 160/220 = 0.73 A
16Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
TV: E = 250 W × 3600 s = 9 × 105 J Toaster: E = 1200 W × 600 s = 7.2 × 105 J
The TV set uses more energy.
17An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Rate of heat = P = I2R = 52 × 44 = 1100 W (1100 J/s)
18Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
Tungsten has a very high melting point (3380 °C) and high resistivity. It can be heated till it glows without melting.
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Alloys have higher resistivity than pure metals, so they produce more heat. They also do not oxidise (burn) readily at high temperatures.
(c) Why is the series arrangement not used for domestic circuits?
In series, all devices get the same current and share the voltage. So each device does not get the full voltage it needs. If one device fails, the circuit breaks and all devices stop working. So series arrangement is not used.
(d) How does the resistance of a wire vary with its area of cross-section?
Resistance is inversely proportional to the area of cross-section. If the area increases, the resistance decreases.
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Copper and aluminium have very low resistivity. So they are good conductors and very little energy is lost as heat. Aluminium is also light and cheap.
