NIOS Class 10 Science and Technology Chapter 16 Light Energy Solutions to each chapter is provided in the list so that you can easily browse through different chapters NIOS Class 10 Science and Technology Chapter 16 Electrical Energy and select need one. NIOS Class 10 Science and Technology Chapter 16 Electrical Energy Question Answers Download PDF. NIOS Study Material of Class 10 Science and Technology Notes Paper 212.
NIOS Class 10 Science and Technology Chapter 16 Electrical Energy
Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 10 Science and Technology Chapter 16 Electrical Energy, NIOS Secondary Course Science and Technology Solutions for All Chapters, You can practice these here.
Electrical Energy
Chapter: 16
INTEXT QUESTIONS 16.1
1. Define the units of
(i) Charge.
Ans: The SI unit of electric charge is the coulomb which is a derived SI unit and is represented by the symbol C. A coulomb is defined as the amount of charge that passes through an electrical conductor carrying one ampere per second.
(ii) Electric potential.
Ans: The electric potential at a point is the amount of work done to move a unit positive charge from an infinitely long distance to that point. The SI unit of electric potential is volt (V), and it can also be written as Joule per Coulomb.
2. When a glass rod is rubbed with a piece of silk it acquires +10 micro coulomb of charge. How many electrons have been transferred from glass to silk?
Ans:
3. How will the force between two small electrified objects vary if the charge on each of the two particles is doubled and separation is halved?
Ans:
4. How does the force between two small charged spheres change if their separation is doubled?
Ans: F = ¼ F.
5. A particle carrying a charge of 1 micro coulomb (μC) is placed at a distance of 50 cm from a fixed charge where it has a potential energy of 10 J. Calculate.
(i) The electric potential at the position of the particle.
Ans:
(ii) The value of the fixed charge.
Ans:
6. Two metallic spheres A and B mounted on two insulated stands as shown in the Fig. 16.4 are given some positive and negative charges respectively. If both the spheres are connected by a metallic wire, what will happen?
Ans: Electrons will flow from sphere B to sphere A through the wire till the potentials of the two spheres become equal.
INTEXT QUESTIONS 16.2
1. Define the SI units of
(i) Current
Ans: Unit of current is ampere. 1A is the current in a wire in which 1C charge flows in 1 second.
(ii) Resistance.
Ans: Unit of resistance is ohm. 1 ohm is the resistance of a wire across which when 1V potential difference is applied, 1A current flows through it.
2. Name the instruments used to measure
(i) Current
Ans: Ammeter.
(ii) potential difference.
Ans: Voltmeter.
3. Why is a conductor different from an insulator?
Ans: A conductor has free electrons, whereas an insulator has no free electrons.
4. How is a volt related to an ohm and an ampere?
Ans: A conductor has free electrons, whereas an insulator has no free electrons.
5. A number of bulbs are connected in a circuit. Decide whether the bulbs are connected in series or in parallel, when
(i) the whole circuit goes off when one bulb is fused.
Ans: If the whole circuit goes off when one bulb is fused, the bulbs are connected in series.
(ii) only the bulb that get fused goes off.
Ans: If any one bulb goes off and the rest of the circuit remains working, the bulbs are connected in parallel.
6. When the potential difference across a wire is doubled, how will the following quantities be affected by
(i) Resistance of the wire.
Ans: Resistance of the wire remains unaffected.
(ii) Current flowing through the wire?
Ans: Current flowing through the wire is doubled.
7. What is the reading of ammeter in the circuit given below?
Ans: 1A.
8. How can three resistors of resistance 2Ω, 3Ω and 6Ω be connected to give a total resistance of
(i) 11Ω
Ans: All the three resistors are connected in series.
(ii) 4.5Ω
Ans: Resistors 2Ω and 6Ω are connected in parallel and 3Ω is connected in series to the combination of 2Ω and 6Ω.
(iii) 4Ω?
Ans: Resistors 3Ω and 6Ω are connected in parallel and 2Ω is connected in series to the combination of 3Ω and 6Ω.
9. State two advantages of connecting electrical devices in parallel with the battery instead of connecting them in series.
Ans: In a parallel circuit, every electrical gadget operates separately because they take current as per their requirement. Total resistance of the circuit is decreased. If one component fails, the circuit is not broken and other electrical devices work properly.
INTEXT QUESTIONS 16.3
1. Which will produce more heat in 1 second – 1 ohm resistance on 10V or a 10 ohm resistance on the same voltage? Give reason for your answer.
Ans: Q/t = V2/R. This implies that more the resistance less the power. Therefore, more heat will flow in 1s in 1 ohm resistor.
2. How will the heat produced in a conductor change in each of the following cases?
(i) The current flowing through the conductor is doubled.
Ans: Heat produced becomes four times.
(ii) Voltage across the conductor is doubled.
Ans: Heat produced becomes four times.
(iii) Time for which current passed is doubled.
Ans: Heat produced will be doubled.
3. 1 A current flows through a conductor of resistance 10 ohms for 1/2 minute. How much heat is produced in the conductor?
Ans: Q = i 2Rt = 1 × 10 × 30 = 300 J.
4. Two electric bulbs of 40 W and 60 W are given. Which one of the bulbs will glow brighter if they are connected to the mains in
(i) Series.
Ans: (i) The bulb with lowest wattage (highest resistance) glows with maximum brightness.
(ii) Parallel?
Ans: (ii) The bulb with highest wattage (lowest resistance) glows with maximum brightness.
5. How is 1 kWh related with SI unit of energy?
Ans: 1 k W h = 3.6 × 106 J.
6. Name two household electric devices based on thermal effect of electric current.
Ans: Two house electric device based on thermal effect of electric current are mentioned below:
(i) Electric heater.
(ii) Electric kettle.
INTEXT QUESTIONS 16.4
1. Which has a higher resistance, a 40W-220 V bulb, or a 1 kW-220V electric heater?
Ans: V2 / P = 40 lamp has higher resistance.
2. What is the maximum current that a 100W, 220 V lamp can withstand?
Ans:
3. How many units of electricity will be consumed by a 60 W lamp in 30 days if the bulb is lighted 4 hours daily?
Ans: Q = Pt = 60W X 4h X 30 = 7200 W h = 7.0 kW.
4. How many joules of electrical energy will a quarter horse power motor consume in one hour?
Ans:
5. An electric heater is used on 220 V supply and draws a current of 5 A. What is its electric power?
Ans: P = VI = 220 V X 5A = 1100 W.
6. Which uses more energy, a television of 250 W in 60 minutes or a toaster of 1.2 kW in (1/6)th of an hour?
Ans: Energy used by television = 0.25 k w x 1 h = kw h.
Energy used by toaster = 1.2 kw x ⅙ = .0.2kw h .
| TERMINAL EXERCISE |
1. Tick mark the most appropriate answer out of four given options at the end of each of the following statements:
(i) Q.
(ii) Q/2.
(iii) Zero.
(iv) 2Q.
Ans: (iii) Zero.
(b) J C–1 is the unit of
(i) Current.
(ii) Charge.
(iii) Resistance.
(iv) Potential.
Ans: (iv) Potential.
(c) Which of the following materials is an electrical insulator?
(i) Mica.
(ii) Copper.
(iii) Tungsten.
(iv) Iron.
Ans: (i) Mica.
(d) The device which converts chemical energy into electrical energy is called
(i) Electric fan.
(ii) Electric generator.
(iii) Electric cell.
(iv) Electric heater.
Ans: (iii) Electric cell.
(e) The resistance of a conductor does not depend on its
(i) Temperature.
(ii) Length.
(iii) Thickness.
(iv) Shape.
Ans: (iv) Shape.
(f) There are four resistors of 12 Ω each. Which of the following values is possible by their combination (series and/or parallel)?
(i) 9 Ω.
(ii) 16 Ω.
(iii) 12 Ω.
(iv) 30 Ω.
Ans: (iv) 30 Ω.
(g) In case of the circuit shown below in Fig. 16.12, which of the following statements is/are true:
(i) R1, R2, and R3 are in series.
(ii) R2 and R3 are in series.
(iii) R2 and R3 are in parallel.
(iv) The equivalent resistance of the circuit is [R1+ (R2 R3/R2 + R3)].
Ans: (iv) The equivalent resistance of the circuit is [R1+ (R2 R3/R2 + R3)].
(h) The equivalent resistance of two resistors of equal resistances connected in parallel is —— the value of each resistor:
(i) Half.
(ii) Twice.
(iii) Same.
(iv) One fourth.
Ans: (i) Half.
2. Fill in the blanks.
(a) When current is passed through a conductor, its temperature _________.
Ans: Increased.
(b) The amount of ____________ flowing past a point per unit ____________ is defined as electric current.
Ans: Electric charge.
(c) A current carrying conductor carries an ____________.
field around it.
Ans: Magnetic.
(d) One ampere equals one ____________. per _________.
Ans: Columb and Second.
(e) Unit of electric power is ____________.
Ans: Watt.
(f) Of the two wires made of the same material and having the same thickness, the longer one has ____________ resistance.
Ans: More.
3. How many types of electric charge exist?
Ans: There are two types of charge exist are mentioned below:
(i) positive.
(ii) Negative.
4. In a nucleus there are several protons, all of which have positive charge. Why does the electrostatic repulsion fail to push the nucleus apart?
Ans: A nucleus of an atom has protons and neutrons. we know that a proton has a charge of + 1, while a neutron has no charge, or 0. Therefore, the nucleus of an atom will always have a positive charge.
Since protons have charge + 1 e, they experience an electric force that tends to push them apart, but at short range the attractive nuclear force is strong enough to overcome the electrostatic force. The nuclear force binds nucleons into atomic nuclei.
5. What does it mean to say that charge is conserved?
Ans: The net electric charge in the system after the charging by rubbing becomes + q + ( − q ) = q − q = 0 . In conclusion, the net charge on the system in the final state is the same as in the initial state, i.e., zero. This constant value of net electric charge proves the law of conservation of electric charge.
6. A point charge of +3.0 μC is 10 cm apart from a second point charge of –1.5 μC. Find the magnitude and direction of force on each charge.
Ans: As we know force, F=kq1q2/r^2;
Here, k = 9×10^9, q1 = +3×10^-6; q2 =-1.5×10^-6. R = 10cm = 0.1m.So, F = 9×10^9× (3×10^-6× -1.5×10^-6 / 0.1)Or, F = 4.05N.
7. Name the quantity measured by the unit (a) VC (b) Cs–1.
Ans: The unit of charge quantity is VC and ampere is measured by Cs^-1.
8. Give a one word name for the unit.
(a) JC–1.
(b) Cs–1.
Ans: The SI unit of Electric potential is Volts(v) or Joules per coulomb (JC-1).
9. What is the potential difference between the terminals of a battery if 250 J of work is required to transfer 20 C of charge from one terminal of the battery to the other?
Ans: We know, Q = CV,
Here, Q = 250J, C = 20c, v = ?
So, 250 = 20 × V,
V = 12.5volt.
10. Give the symbols of
(a) Cell.
Ans: Do yourself.
(b) Battery.
Ans:
(c) resistor.
Ans:
(d) voltmeter.
Ans:
11. What is the conventional direction of flow of electric current? Do the charge carriers in the conductor flow in the same direction? Explain.
Ans: The conventional flow of current is in the opposite direction of electrons, i.e. from the positive terminal to the negative terminal of the cell.
The charge carriers in a conductor, such as electrons in a metal wire, actually flow in the opposite direction, from the negative terminal to the positive terminal. This is because electrons are negatively charged particles and are repelled by the negative terminal and attracted towards the positive terminal. The convention of current flow from positive to negative is a historical artefact dating back to the time when it was not yet known that electrons were the primary charge carriers. It was initially believed that electric current was the flow of positive charges, hence the convention. This convention has persisted for consistency and convenience in calculations and circuit analysis, even though it doesn’t reflect the actual movement of charge carriers.
12. Out of ammeter and voltmeter which is connected in series and which is connected in parallel in an electric circuit?
Ans: They are connected in series whereas the voltmeter is connected in parallel in the circuit of the electrical system.
13. You are given two resistors of 3 Ω and 6 Ω, respectively. Combining these two resistors what other resistances can you obtain?
Ans: If two resistors of 3 Ω and 6 Ω, respectively are combined in a circuit, the various values of final resistance that can be obtained are 9 Ω and 2 Ω.
14. What is the current in SI unit if+100 coulombs of charge flows past a point every five seconds?
Ans: As we know, current(i) = Q/t;
So, current = 100/5 = 20A.
Then 20A current will transfer through the circuit.
15. Deduce an expression for the electrical energy spent in flow of current through a conductor.
Ans: The expression for the electrical energy spent in the flow of current through an electrical appliance is: E = VIt = I2Rt.
16. Find the value of resistor X as shown in Fig. 16.13.
Ans: Do your self.
17. In the circuit shown in Fig. 16.14, find
(i) Total resistance of the circuit.
(ii) Ammeter reading.
(iii) Current flowing through 3 Ω resistor.
Ans: Do your self.
18. For the circuit shown in Fig. 16.15, find the value of:
(i) Current through 12Ω resistor.
(ii) Potential difference across 6Ω and 18Ω resistors.
Ans: Do your self.
19. You are given three resistors of 1 Ω, 2 Ω and 3 Ω. Show by diagrams, how will you connect these resistors to get:
(a) 6/11 Ω.
(b) 6 Ω.
(c) 1.5 Ω.
Ans: Do yourself.
20. A resistor of 8 Ω is connected in parallel with another resistor of X Ω. The resultant resistance of the combination is 4.8 ohm. What is the value of resistor X?
Ans: The equation will be,
1/8 + 1/X = 1/4.8;
Or, (8+ X)/ 8X = 1/4.8;
Or, 3.2X= 38.4;
So, X= 38.4/3.2 = 12 ohm.
21. In the circuit Fig. 16.16, find
(i) Total resistance of the circuit.
(ii) Total current flowing through the circuit.
(iii) The potential difference across 4 Ω resistor.
Ans: Do yourself.
22. How many 132 Ω resistors should be connected in parallel to carry 5 A current in 220 V line?
Ans: Here resistance, R = 220 / 5 = 44 ohm,
Let the number of resistors is n.
By equation, 132/n = 44; or, n = 3.
