CBSE Class 10 Science Electricity VBQs Set 06

These value based and competency based questions on Class 10 Science Chapter Electricity connect Ohm's law, resistivity, series and parallel circuits and electrical power to everyday life: household wiring, fuses, earthing, electricity bills and saving energy. Every question has a full worked answer, a simple-words summary and a teacher's note.

Case Based Questions

Question 1. Read the following case based passage and answer the questions given after the passage.
A room has two tube lights, a fan and a TV. Each tube light draws 40 W, the fan draws 80 W and the TV draws 60 W on the average. The tube lights are kept on for five hours, the fan for twelve hours and the TV for eight hours every day. The rate for electrical energy is ₹ 3.10 per kWh.
(i) Calculate the energy consumed by each tube light in a day.
(ii) What is the total energy consumed in a day?
(iii) Find the cost of electricity used in this room in a 30-day month.
(iv) What quantities determine the rate at which energy is delivered by current?
Answer:
(i) E = P × t = 0.04 kW × 5 h = 0.2 kWh.
(ii) Fan: 0.08 × 12 = 0.96 kWh; TV: 0.06 × 8 = 0.48 kWh; tube lights: 2 × 0.2 = 0.4 kWh. Total = 0.4 + 0.96 + 0.48 = 1.84 kWh.
(iii) Energy in 30 days = 30 × 1.84 = 55.2 kWh. Cost = 55.2 × 3.10 = ₹ 171.12.
(iv) The rate of delivery of energy (power, P = VI) depends on the potential difference across the conductor and the current flowing in the circuit.
In simple words: Energy in kWh is power in kW times hours of use.

Teacher's Note:
1. Convert W to kW before multiplying by hours.
2. Cost = units consumed × rate per unit.

Question 2. Read the following passage and answer the questions that follow.
Shivani had studied in her class that an electric bulb consists of a filament made up of a metal having high melting point such as tungsten. Also, the bulbs are filled with a chemically inert gas such as nitrogen or argon to prolong the life of filament.
(A) A current of 2 A is drawn by a filament of an electric bulb. What will be the number of electrons passing through a cross section of the filament in 32 seconds?
(B) Shivani wondered how can electricity produce heat? Explain.
(C) What is the meaning of the following statement? ‘A typical rating on an electric bulb is 60 W–220 V.’
Answer:
(A) Q = It = 2 × 32 = 64 C. Number of electrons n = Q/e = 64 / (1.6 × 10⁻¹⁹) = 4 × 10²⁰.
(B) When a potential difference is applied across a conductor, free electrons move through it but collide with the atoms of the conductor. In these collisions the electrons transfer a part of their energy to the atoms, which vibrate more vigorously, so the temperature of the conductor rises. In this way electrical energy is converted into heat.
(C) The bulb works best at a potential difference of 220 V, and at this voltage it consumes electrical energy at the rate of 60 W (60 J every second).
In simple words: Moving electrons collide with atoms and heat the wire; the bulb label gives its power and working voltage.

Teacher's Note:
1. Use Q = It and n = Q/e in part (A).
2. Part (B) is the idea behind Joule heating.

Teacher's Note:
1. Compare the heaters at the same voltage.
2. Change mA to A before using P = VI.

Question 4. Read the following passage and answer the questions.
Ohm’s law is used to maintain the desired voltage drop across the electronic components. We can control the speed of the fans at our homes by moving the regulator to and fro. Here the current flowing through the fan is controlled by regulating the resistance through the regulator. A circular knob on the component can be rotated to achieve a variable resistance on the output terminals.
(A) In which condition for a fixed supply of voltage, the current flowing through a conductor will decrease?
(B) A resistance of 5 ohms is further drawn so that its length becomes double. Calculate its new resistance.
(C) How would you control the current flowing through a fan?
Answer:
(A) For a fixed voltage, I = V/R, so the current decreases when the resistance increases, e.g. when the length of the conductor increases (or its area decreases).
(B) When the wire is drawn to double the length, its area becomes half, so R′ = ρ(2l)/(A/2) = 4R = 4 × 5 = 20 Ω.
(C) By a regulator, which is a variable resistor (rheostat). Rotating the knob changes the length of resistance wire in the circuit, which changes the resistance and hence the current through the fan.
In simple words: A fan regulator is a variable resistor that changes the current.

Teacher's Note:
1. Link current and resistance with Ohm's law.
2. Volume stays constant when a wire is drawn.

Question 5. Read the following case based passage and answer the questions given after the passage.
In household electric circuits, the mains supply is delivered to our homes using a three core cable. The cable consists of three wires: live wire, neutral wire and earth wire. The potential difference between the live and neutral wire is 220 volts. The live wire is connected to the electric meter through a fuse or a circuit breaker of higher rating. The neutral wire is connected directly to the electric meter.
(i) What is a short circuit?
(ii) How are switches connected in the circuit?
(iii) What is the usual current rating of the fuse wire in the line if an electric iron, geyser, room heater, etc. are in use?
(iv) Why is earthing of all electrical appliances recommended?
Answer:
(i) A short circuit occurs when the live wire comes in direct contact with the neutral wire, so that the resistance becomes very small and a very high current flows.
(ii) Switches are connected in the live wire, so that when a switch is off no part of the appliance remains at a high potential (220 V).
(iii) A fuse of rating 15 A is usually used for appliances of high power such as an electric iron, geyser and room heater.
(iv) Earthing protects the user from an electric shock by taking away any leakage current to the ground through the metallic body of the appliance.
In simple words: Fuse and switch go on the live wire, and earthing keeps users safe.

Teacher's Note:
1. Fuses of 5 A are used for lights and fans, 15 A for heavy appliances.
2. The earth wire is connected to the metal body of the appliance.

Question 6. Read the following case based passage and answer the questions given after the passage.
Resistance of a conductor depends on the length, area of cross-section and nature of the material of the conductor. When a conductor is stretched (increased in its length), its area of cross-section decreases accordingly but the volume (area × length) of the conductor remains the same. Resistivity of the conductor ρ = RA/l.
(i) What do you mean by resistivity?
(ii) The SI unit of resistivity is (a) ohm (b) ohm-metre (c) ohm²-metre (d) ohm-metre².
(iii) Write one difference between resistance and resistivity.
(iv) If the length of a wire of resistance R is halved and its area of cross-section is doubled, its new resistance R′ will be (a) R (b) R/2 (c) R/4 (d) R/8.
Answer:
(i) Resistivity is the resistance of a conductor of unit length and unit area of cross-section.
(ii) (b) ohm-metre, since ρ = RA/l has the unit Ω m²/m = Ω m.
(iii) Resistance is the property of a conductor that opposes the flow of charge; it depends on the dimensions and its unit is Ω. Resistivity is a characteristic property of the material, which changes only with temperature; its unit is Ω m.
(iv) R′ = ρ(l/2)/(2A) = (1/4)(ρl/A) = R/4, so (c).
In simple words: Resistance depends on shape; resistivity depends only on the material.

Teacher's Note:
1. Derive the SI unit from the formula.
2. Combine the length factor and the area factor.

Question 7. Read the following case based passage and answer the questions given after the passage.
An electrician is a tradesman specializing in electrical wiring of buildings, transmission lines, stationary machines and related equipment. An electrician may be employed in the installation of new electrical components or the maintenance and repair of existing electrical infrastructure. An electrician has made the electric circuit of a house in such a way that if a fan is closed, the lamps also stop glowing.
(i) What is the defect in this type of circuit wiring? Why?
(ii) Two resistances R₁ and R₂ are connected turn by turn in parallel and in series. In which case will the resultant resistance be less than either of the individual resistances?
(iii) Which is the better way to connect lights and other appliances in a domestic circuit: series connection or parallel connection?
(iv) In which type of combination will different resistors have equal value of electric current through them?
Answer:
(i) The electrician has connected the appliances in series. In series, if one device is switched off or fails, the circuit is broken and all the other devices stop working.
(ii) In parallel, because R_p = R₁R₂/(R₁ + R₂) is less than either, whereas in series R_s = R₁ + R₂ is greater than either.
(iii) Parallel connection is better, because each device gets the full potential difference and keeps working even if other devices stop.
(iv) In a series combination, all resistors carry the same current.
In simple words: Home wiring is parallel so every appliance works independently.

Teacher's Note:
1. Series: same current. Parallel: same voltage.
2. Mention the independent on/off control of each appliance.

Question 8. Read the following case based passage and answer the questions given after the passage.
Every electrical appliance like an electric bulb, radio or fan has a label which tells us the voltage to be applied and the electrical power consumed by it. For example, a bulb marked 100 W–220 V has a power consumption of 100 watts when used on 220 volts. The power rating tells us the rate at which electrical energy is consumed, and the current drawn is found from P = V × I. Usual ratings: tube light 40 W, electric bulb 60 W, radio set 80 W, electric fan 100 W, T.V. set 120 W, refrigerator 150 W, electric iron 750 W, electric heater 1000 W, immersion heater 1500 W, washing machine 3000 W (all at 220 V).
(i) What is the current drawn by an electric fan?
(ii) An electric heater is used for 2 h at 220 V. What is the energy consumed?
(iii) What is the resistance of an electric bulb of 60 W?
(iv) What is the net current drawn by a parallel combination of a tube light, an electric bulb and a radio set when connected to a 220 V power supply?
Answer:
(i) I = P/V = 100/220 = 0.45 A.
(ii) E = P × t = 1 kW × 2 h = 2 kWh.
(iii) R = V²/P = 220²/60 ≈ 807 Ω.
(iv) Total power = 40 + 60 + 80 = 180 W, so I = 180/220 = 0.82 A.
In simple words: Current drawn = rated power divided by supply voltage.

Teacher's Note:
1. Parallel: currents add, or just add the powers.
2. Remember the supply voltage is 220 V.

Question 9. Read the following case based passage and answer the questions given after the passage.
The resistivity of alloys is much higher than that of the pure metals from which they are made. For example, the resistivity of manganin (an alloy of copper, manganese and nickel) is about 25 times that of copper, and the resistivity of constantan (an alloy of copper and nickel) is about 30 times that of copper. It is due to their high resistivities that manganin and constantan are used to make resistance wires used in electronic appliances to reduce the current in a circuit. Another alloy having a high resistivity is nichrome, an alloy of nickel, chromium, manganese and iron, with a resistivity about 60 times that of copper. The heating elements of electrical heating appliances such as electric iron and toaster are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher and an alloy does not undergo oxidation (burn) easily even at high temperature.
(i) What is the use of metals which have less resistance?
(ii) What are the applications of metals which have high resistance?
(iii) Nichrome alloy is used for making the heating elements of electrical appliances such as heaters (geysers) and hair dryers. Why?
(iv) Name the alloys which have very high resistivity as compared to copper.
Answer:
(i) Metals of low resistance are used as conducting wires.
(ii) Metals or alloys of high resistance are used for heating elements and resistors.
(iii) Nichrome has a very high resistivity, so the element has a high resistance and produces a lot of heat on passing current. It also does not oxidise (burn) easily even at high temperature, so it can be kept red hot in air without breaking.
(iv) Manganin, constantan and nichrome.
In simple words: High-resistivity alloys heat up strongly and do not burn out.

Teacher's Note:
1. Mention two properties in (iii): high resistivity and no oxidation.
2. Copper and aluminium are used for connecting wires.

Teacher's Note:
1. Estimate the largest current and voltage before choosing the meters.
2. The source key for (iv) was (c); it is corrected to (b) because A₂ cannot carry 2.25 A.

Question 11. In the given circuit, A, B, C and D are four lamps connected with a battery of 60 V (the currents in the lamps are 3 A, 4 A, 5 A and 3 A respectively). Analyse the circuit to answer the following questions. (i) What kind of combination are the lamps arranged in (series or parallel)? (ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps? (iii) Explain with proper calculations which lamp glows the brightest. (iv) Find out the total resistance of the circuit.

Answer:
(i) All lamps have the same voltage (60 V) but different currents, so they are in parallel.
(ii) (a) If one lamp is faulty it does not affect the working of the others. (b) Each lamp uses the full potential of the battery.
(iii) P = VI: A = 60 × 3 = 180 W; B = 60 × 4 = 240 W; C = 60 × 5 = 300 W; D = 60 × 3 = 180 W. Lamp C has the greatest power, so it glows the brightest.
(iv) Total current I = 3 + 4 + 5 + 3 = 15 A, so R = V/I = 60/15 = 4 Ω.
In simple words: In parallel the lamp with the largest current has the greatest power.

Teacher's Note:
1. Brightness depends on power, P = VI.
2. Total current is the sum of the branch currents.

Question 12. Read the following and answer the questions.
A cell or a battery is the source of electrical energy. Due to the chemical reactions inside them a potential difference is set up which is responsible for the flow of current through any electrical circuit. To maintain this flow, the source continuously has to provide the energy. Only a part of this energy is used in useful work; the rest may be consumed in the form of heat by raising the temperature of the appliances.
(i) How is the heat produced in a resistor related to the current flowing in that resistor?
(ii) Give two practical applications of heating effect of current.
(iii) Which type of energy is transformed into heat energy?
(iv) An electric iron of resistance 25 Ω takes a current of 7 A. Calculate the heat developed in 0.5 min.
(v) 200 J of heat is produced in 10 s in a 5 Ω resistance. Find the potential difference across the resistor.
Answer:
(i) The heat produced is directly proportional to the square of the current: H = I²Rt.
(ii) Electric heater, electric iron, toaster, electric bulb and fuse (any two).
(iii) Electrical energy is transformed into heat energy.
(iv) H = I²Rt = 7² × 25 × 30 = 36 750 J = 36.75 kJ.
(v) Power P = H/t = 200/10 = 20 W = I²R, so I² = 20/5 = 4 and I = 2 A. Then V = IR = 2 × 5 = 10 V.
In simple words: Heat = I²Rt; 200 J in 10 s in 5 Ω needs 2 A and 10 V.

Teacher's Note:
1. Convert minutes to seconds.
2. First find the power, then the current.

Question 13. Read the following passage and answer the questions that follow.
Make a parallel combination, XY, of three resistors having resistances R₁, R₂ and R₃. Connect it with a battery, a plug key and an ammeter. Also connect a voltmeter in parallel with the combination of resistors. Let us take resistances R₁, R₂ and R₃ of values 2 Ω, 4 Ω and 12 Ω respectively and let V = 6 V. Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. Then insert the ammeter in series with the resistor R₁ and note the ammeter reading I₁. Similarly, measure the currents through R₂ and R₃, say I₂ and I₃.
(A) The voltmeter and ammeter readings (in the first set-up) are: (a) 6 V, 7.2 A (b) 6 V, 5 A (c) 3 V, 10 A (d) 3 V, 1.5 A. Select the correct row.
(B) The reading of the ammeter when it is connected in series with the resistor R₁ is: (a) 0.5 A (b) 1.5 A (c) 3 A (d) 5 A.
(C) What is the relation between the total current and the currents flowing in each of the resistors?
(D) What is the relation between the current flowing in resistors R₂ and R₃ and their resistances?
Answer:
(A) 1/R_p = 1/2 + 1/4 + 1/12 = 10/12, so R_p = 1.2 Ω and I = 6/1.2 = 5 A. The voltmeter reads 6 V: (b).
(B) I₁ = V/R₁ = 6/2 = 3 A: (c).
(C) The total current is the sum of the branch currents: I = I₁ + I₂ + I₃ (5 = 3 + 1.5 + 0.5).
(D) I₂ = 6/4 = 1.5 A and I₃ = 6/12 = 0.5 A, so I₂ : I₃ = 3 : 1 while R₂ : R₃ = 1 : 3. The currents are inversely proportional to the resistances.
In simple words: In parallel the same voltage drives different currents, in inverse proportion to resistance.

Teacher's Note:
1. Find R_p first and then the total current.
2. Verify with I = I₁ + I₂ + I₃.

Value Based and Real Life Questions

Question 14. You have been assigned a duty to create awareness in your school about saving electricity. (i) Write any two ways by which you will create awareness among your schoolmates about saving electricity. (ii) Explain how saving electricity is important at individual level as well as at national level.
Answer:
(i) (a) Speak about it in the morning assembly in school. (b) Put posters on saving electricity on the school notice board (and ask everyone to switch off lights and fans when not needed, and to use LED bulbs).
(ii) At the individual level, saving electricity reduces the electricity bill and the waste of energy.
At the national level, the electricity saved can be supplied to villages that do not have electricity and used in industries, agriculture and other useful purposes; it also reduces the burning of fossil fuels and so helps the national economy and the environment.
In simple words: Saving electricity saves money for us and gives power to those who lack it.

Teacher's Note:
1. Give practical and realistic steps.
2. Link individual saving with the national benefit.

Question 15. Why do we connect a large number of electric bulbs for decorating buildings as during festivals such as Diwali or marriage function in series circuit and not in parallel circuit?
Answer:
The small decorative bulbs are made to work at a low voltage (a few volts). When a large number of them are connected in series, the 220 V of the mains gets divided among them, so each bulb gets its rated voltage and does not fuse.
The series arrangement also needs only a single pair of connecting wires. (The disadvantage is that if one bulb fuses, the whole string goes off.)
In simple words: In series the mains voltage is shared by many low-voltage bulbs.

Teacher's Note:
1. Each bulb gets 220 V divided by the number of bulbs.
2. Mention the drawback of the series string too.

Question 16. Rita designed a circuit in which resistors are connected in series. Yet she is not satisfied with the series resistors because if there is a fault in some component of the circuit, the whole circuit stops working. What would be your suggestion in alternative to the resistance in series?
Answer:
I would suggest connecting the resistors in parallel. In a parallel connection each component has its own path, so if one component is faulty the current continues to flow through the other components and they keep working.
In simple words: Parallel wiring keeps the other branches working when one fails.

Teacher's Note:
1. Each branch works independently in parallel.
2. This is how house wiring is done.

Question 17. Explain the role of fuse wire connected in series with any electrical appliance in an electric circuit. Why should a fuse with a defined rating for an electric circuit not be replaced by the one with a larger rating?
Answer:
A fuse wire is a safety device connected in series with the live wire. It has a low melting point, so when a large current flows (overloading or short circuit) it melts and disconnects the appliance from the supply.
If a fuse of a larger rating is used instead, a current larger than the safe value can pass without melting it, so the circuit and the appliance will not be protected and may be damaged or catch fire.
In simple words: A fuse of the correct rating blows before the wires overheat.

Teacher's Note:
1. Always replace a fuse by one of the same rating.
2. A fuse is a weak link in the circuit.

Question 18. Why do we get electric shock in damp conditions?
Answer:
Water provides a conducting path for the current to flow through the human body in damp conditions such as a bathroom. A wet body has a low resistance, so a large current can easily pass through it, leading to an electric shock.
In simple words: Wet skin has low resistance, so a dangerous current flows.

Teacher's Note:
1. Never touch switches with wet hands.
2. Use earthing and insulated footwear.

Question 19. What will happen when the live and the neutral wires in a circuit touch each other due to defective wiring?
Answer:
The resistance of the circuit becomes very small, so a very large current flows. This is called a short circuit. A large amount of heat is produced, which can cause a fire; the fuse should melt and break the circuit.
In simple words: Touching live and neutral wires causes a short circuit.

Teacher's Note:
1. Link it with a fuse.
2. Large current means large heat.

Question 20. Why do we do the earthing of an electrical appliance?
Answer:
The earth wire connects the metal body of the appliance to the ground. If there is a leakage of current to the body, it flows to the earth through the low-resistance earth wire, so the user is protected from a severe electric shock.
In simple words: Earthing carries leakage current safely to the ground.

Teacher's Note:
1. Earth wire is connected to the metal body.
2. Earth pin is longer and thicker.

Question 21. Why do electricians wear rubber hand gloves while working?
Answer:
Rubber is an electrical insulator, so it does not allow current to pass through the body. The electrician can work on a circuit without a risk of getting an electric shock.
In simple words: Rubber gloves insulate the electrician from the current.

Teacher's Note:
1. Insulators have very high resistance.
2. Dry rubber is a safe insulator at household voltages.

Question 22. Why is lead-tin alloy used for fuse wires?
Answer:
The lead-tin alloy has a low melting point. When the current exceeds the safe limit it melts quickly and breaks the circuit, preventing fire and damage.
In simple words: A fuse wire must melt easily when the current is too large.

Teacher's Note:
1. Fuse wire has a low melting point and a suitable resistance.
2. It is connected in series.

Question 23. Why should the melting point of a fuse wire be low?
Answer:
A fuse wire must melt quickly as soon as the current exceeds the safe limit, to break the circuit and protect the appliances. This is possible only if its melting point is low.
In simple words: A fuse is a deliberate weak link that melts first.

Teacher's Note:
1. Use a lead-tin alloy for fuse wire.
2. It is connected in series.