Answer:
An electric circuit is a continuous and closed path through which electric current flows. It consists of a source of electricity (like a cell or battery), conducting wires, and electrical components (like a bulb or resistor). The circuit must be closed for current to flow; if broken anywhere (e.g., by opening a switch), the current stops. (Page 172, Section 11.1)
2Define the unit of current.
Answer:
The SI unit of electric current is ampere (A). One ampere is defined as the flow of one coulomb of charge per second. That is, 1 A = 1 C/1 s. (Page 172, Section 11.1)
3Calculate the number of electrons constituting one coulomb of charge.
Answer:
We know that charge on one electron = 1.6 × 10-19 C.
Number of electrons in 1 coulomb = 1 / (1.6 × 10-19) = 6.25 × 1018 electrons. (Page 172, Section 11.1)
4Name a device that helps to maintain a potential difference across a conductor.
Answer:
A battery or a cell helps to maintain a potential difference across a conductor. (Page 173, Section 11.2)
5What is meant by saying that the potential difference between two points is 1 V?
Answer:
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of charge from one point to the other. (Page 173, Section 11.2)
6How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Energy given to each coulomb of charge = Potential difference = 6 J/C.
Therefore, 6 joules of energy is given to each coulomb of charge passing through a 6 V battery. (Page 174, Section 11.2)
7On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends on:
Its length (directly proportional)
Its area of cross-section (inversely proportional)
The nature of its material (resistivity)
Temperature
(Page 178, Section 11.5)
8Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
Current will flow more easily through a thick wire because resistance is inversely proportional to the area of cross-section. A thick wire has a larger cross-sectional area, hence lower resistance, allowing more current to flow. (Page 178, Section 11.5)
9Let 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?
Answer:
According to Ohm's law (I = V/R), if resistance remains constant and potential difference is halved, the current will also be halved. (Page 176, Section 11.4)
10Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
Coils of electric toasters and irons are made of alloys because:
Alloys have higher resistivity than pure metals, producing more heat.
Alloys do not oxidize (burn) readily at high temperatures.
(Page 179, Section 11.5)
11Use 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?
Answer:
(a) Iron is a better conductor than mercury because iron has lower resistivity (10.0 × 10-8 Ω m) compared to mercury (94.0 × 10-8 Ω m).
(b) Silver is the best conductor among the listed materials as it has the lowest resistivity (1.60 × 10-8 Ω m).
Material
Resistivity (Ω m)
Silver
1.60 × 10-8
Copper
1.62 × 10-8
Aluminium
2.63 × 10-8
Iron
10.0 × 10-8
Mercury
94.0 × 10-8
(Page 179, Table 11.2)
12Judge 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.
Answer:
(a) For 1 Ω and 106 Ω in parallel:
1/R = 1/1 + 1/106 = 1 + 0.000001 ≈ 1
Therefore, R ≈ 1 Ω
(b) For 1 Ω, 103 Ω, and 106 Ω in parallel:
1/R = 1/1 + 1/103 + 1/106 = 1 + 0.001 + 0.000001 ≈ 1.001
Therefore, R ≈ 1/1.001 ≈ 0.999 Ω (Page 188, Section 11.6.2)
13An 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?
Answer:
First, find equivalent resistance of the parallel combination:
1/R = 1/100 + 1/50 + 1/500 = (5 + 10 + 1)/500 = 16/500
R = 500/16 = 31.25 Ω
Total current I = V/R = 220/31.25 = 7.04 A
Therefore, the electric iron should have resistance 31.25 Ω and will draw a current of 7.04 A. (Page 188, Section 11.6.2)
14What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
Advantages of parallel connection:
Each device gets the same voltage as the battery.
Devices can be operated independently (switched on/off without affecting others).
If one device fails, others continue to work.
Total resistance decreases, allowing more current to flow as needed by different devices.
(Page 187, Section 11.6.2)
15How 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?
Answer:
(a) For 4 Ω: Connect 3 Ω and 6 Ω in parallel, then connect this combination in series with 2 Ω.
Resistance of parallel combination = (3×6)/(3+6) = 18/9 = 2 Ω
Total resistance = 2 Ω + 2 Ω = 4 Ω
(b) For 1 Ω: Connect all three resistors in parallel.
1/R = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 6/6 = 1
R = 1 Ω (Page 188, Section 11.6.2)
16What 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?
Answer:
(a) Highest resistance: Connect all in series.
R = 4 + 8 + 12 + 24 = 48 Ω
17A 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:
Each part has resistance R/5.
When connected in parallel: 1/R' = 5/(R/5) = 25/R
Therefore, R' = R/25
Ratio R/R' = R/(R/25) = 25
Correct option: (d) 25 (Page 193, Exercise 1)
18Which of the following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d) V²/R
Answer:
Electrical power is given by P = VI = I²R = V²/R.
IR² does not represent electrical power.
Correct option: (b) IR² (Page 193, Exercise 2)
19An 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:
First, find resistance of bulb: R = V²/P = (220)²/100 = 484 Ω
Power at 110 V: P = V²/R = (110)²/484 = 12100/484 = 25 W
Correct option: (d) 25 W (Page 193, Exercise 3)
20Two 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:
Let resistance of each wire be R.
Series combination: Rs = 2R
Parallel combination: Rp = R/2
Heat produced H ∝ 1/R (since H = V²t/R and V, t are constant)
Hs/Hp = Rp/Rs = (R/2)/(2R) = 1/4
Ratio = 1:4
Correct option: (c) 1:4 (Page 193, Exercise 4)
21How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is always connected in parallel across the points between which the potential difference is to be measured. (Page 173, Section 11.2)
22A 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?
Answer:
Given: ρ = 1.6 × 10-8 Ω m, R = 10 Ω, d = 0.5 mm = 0.0005 m
Radius r = d/2 = 0.00025 m
Area A = πr² = 3.14 × (0.00025)² = 1.9625 × 10-7 m²
R = ρl/A → l = RA/ρ = (10 × 1.9625 × 10-7)/(1.6 × 10-8) = 122.66 m
If diameter is doubled, area becomes 4 times.
New resistance R' = R/4 = 10/4 = 2.5 Ω (Page 193, Exercise 6)