CBSE Class 10 Science Electricity Sure Shot Questions Set 02

A second set of 25 sure shot board-style questions from Class 10 Science Electricity, covering Ohm's law, factors affecting resistance, series and parallel combinations, Joule's law, electric power and fuses. Each answer is written step by step with the formula first, followed by a simple-words line and a teacher's note.

Short Answer Questions

Question 1. [2 marks] Explain two disadvantages of series arrangement for household circuit.
Answer:
(i) The current is the same in all appliances in series, so it is impractical to connect a bulb and an electric heater in series, as they need very different currents.
(ii) If one component fails, the circuit is broken and none of the other components works. (They also cannot be switched on and off separately.)
In simple words: In series, one fault switches everything off.

Teacher's Note:
1. Compare with parallel wiring used in homes.
2. Mention both the current and the break-down problems.

Question 2. [2 marks] (i) Why are electric bulbs filled with chemically inactive nitrogen or argon? (ii) What is meant by the statement that the rating of a fuse in a circuit is 5 A?
Answer:
(i) The inert gas prevents the tungsten filament from oxidising and so prolongs the life of the filament.
(ii) It means the maximum current that can safely flow in the circuit is 5 A. If the current exceeds 5 A, the fuse wire melts and breaks the circuit.
In simple words: Inert gas protects the filament; a 5 A fuse blows above 5 A.

Teacher's Note:
1. Remember the fuse is connected in series with the live wire.
2. Choose a fuse rating just above the normal current.

Question 3. [2 marks] (a) What material is used in making the filament of an electric bulb? (b) Name the characteristics which make it suitable for this.
Answer:
(a) Tungsten is used for the filament.
(b) Tungsten has a very high melting point (3380 °C) and a high resistivity, so it can become white hot and give light without melting.
In simple words: Tungsten glows white hot without melting.

Teacher's Note:
1. High melting point plus high resistivity is the pair to remember.
2. The bulb is filled with inert gas to prevent oxidation.

Question 4. [3 marks] Two resistors with resistance 5 Ohm and 10 Ohm respectively are to be connected to a battery of 6 V so as to obtain minimum and maximum current flowing. (A) How will you connect the resistors in each case? (B) Calculate the strength of total current in the circuit in the two cases.
Answer:
(A) For the minimum current the resistance must be maximum, so connect them in series. For the maximum current connect them in parallel.
(B) Series: R = 5 + 10 = 15 Ω, I = 6/15 = 0.4 A.
Parallel: R = (5 × 10)/15 = 3.33 Ω, I = 6/3.33 = 1.8 A.
In simple words: Series gives 0.4 A; parallel gives 1.8 A.

Teacher's Note:
1. Use I = V/R in each case.
2. Maximum current means minimum resistance.

Question 5. [3 marks] Explain in brief the function of an electric fuse in a domestic circuit. An electric heater of current rating 3 kW; 220 V is to be operated in an electric circuit of rating 5 A. What is likely to happen when the heater is switched ‘ON’? Justify your answer with necessary calculation.
Answer:
A fuse is a safeguard that makes sure the current in a domestic circuit does not exceed a safe limit. It melts and breaks the circuit in case of overloading or a short circuit, and so prevents damage to the appliances.
Current drawn by the heater: I = P/V = 3000/220 = 13.6 A.
This is much more than 5 A, so the heater will make the fuse melt (blow) and break the circuit as soon as it is switched on.
In simple words: A 3 kW heater needs 13.6 A, so it blows a 5 A fuse.

Teacher's Note:
1. Compare the current drawn with the circuit rating.
2. A fuse is connected in series in the live wire.

Question 6. [3 marks] Why is a series arrangement not used for domestic circuits?
Answer:
1. The voltage of the source gets divided among the appliances in the ratio of their resistances, so each appliance does not operate at its rated voltage.
2. The total resistance increases and the current decreases, so each appliance gets less power.
3. If any one appliance is switched off or fails, none of the other appliances connected in series with it will operate.
In simple words: In series, the supply is shared and one fault stops everything.

Teacher's Note:
1. Compare with parallel connection.
2. Give at least two reasons for 2 marks.

Teacher's Note:
1. Learn the three points as a table.
2. Metallic wires at constant temperature are ohmic.

Teacher's Note:
1. Learn this table; it is asked often.
2. Use R_s = ΣR and 1/R_p = Σ(1/R).

Question 9. [2 marks] Mention two reasons why tungsten is used for making filaments of electric lamps.
Answer:
(i) Tungsten has a high resistivity, so the filament becomes very hot and gives light.
(ii) Tungsten has a high melting point (3380 °C), so it does not melt at the working temperature.
In simple words: High resistivity makes it glow; the high melting point keeps it from melting.

Teacher's Note:
1. Give two separate reasons.
2. The bulb is also filled with an inert gas.

Question 10. [2 marks] State the difference between a wire used in the element of electric heater and in fuse wire.
Answer:
The wire used in the heating element has a high melting point (and high resistivity), whereas a fuse wire has a low melting point so that it melts when the current is too large.
In simple words: A heater wire must not melt; a fuse wire must melt.

Teacher's Note:
1. Compare the melting points.
2. Heater element: nichrome. Fuse wire: lead-tin alloy.

Question 11. [3 marks] Explain the use of an electric fuse. What type of material is used for fuse wire and why?
Answer:
An electric fuse is used to protect electric circuits and appliances from damage due to overloading and short circuit. It is connected in series with the live wire.
The fuse wire is made of an alloy of low melting point (such as lead and tin) with suitable resistivity, so that it melts quickly when the current exceeds the rated value and breaks the circuit.
In simple words: A fuse melts to cut off the current when it is too large.

Teacher's Note:
1. Name the material and the reason.
2. Fuse in series with the live wire.

Question 12. [3 marks] While studying the dependence of potential difference (V) across a resistor on the current (I) passing through it, in order to determine the resistance of the resistor, a student took 5 readings for different values of current and plotted a graph between V and I. He got a straight line graph passing through the origin. What does the straight line signify? Write the method of determining resistance of the resistor using this graph.
Answer:
The straight line through the origin shows that the potential difference is directly proportional to the current (V ∝ I), i.e. the resistor obeys Ohm's law.
Method: take two points on the line, find ΔV and ΔI between them; the slope ΔV/ΔI of the line gives the resistance R.
In simple words: The slope of a V–I line is the resistance.

Teacher's Note:
1. Choose points far apart on the line.
2. R = V/I = slope.

Question 13. [3 marks] Define resistance. What are the factors on which it depends?
Answer:
Resistance is the property of a conductor to oppose the flow of charges through it; R = V/I, and its SI unit is the ohm (Ω).
It depends on (i) the length of the conductor (R ∝ l), (ii) its area of cross-section (R ∝ 1/A), (iii) the nature of its material (resistivity ρ), and (iv) the temperature: R = ρl/A.
In simple words: Resistance grows with length and temperature and falls with area.

Teacher's Note:
1. Write the formula R = ρl/A.
2. Mention four factors for full marks.

Long Answer Questions

Question 14. [3 marks] State Ohm’s law. How can it be verified experimentally? Does it hold good under all conditions? Comment.
Answer:
Statement: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends: I ∝ V, or V = IR.
Verification: Connect a cell, key, ammeter, the conductor and a voltmeter across it. Vary the voltage and note V and I. The V–I graph is a straight line through the origin, which verifies the law.
Conditions: No, it holds only when the temperature and other physical conditions stay constant (it is not obeyed by devices like diodes).
In simple words: Ohm's law says V/I stays constant for a metal wire at constant temperature.

Teacher's Note:
1. Give the circuit, the graph and the condition to earn full marks.
2. The slope of the V–I graph gives R.

Question 15. [3 marks] State the factors on which the resistance of a cylindrical conductor depends. How will resistance of a conductor change if it is stretched so that its length is doubled?
Answer:
The resistance of a cylindrical conductor is directly proportional to its length l and inversely proportional to its area of cross-section A, and depends on the material: R = ρl/A.
On stretching, the volume remains constant: A′l′ = Al, so with l′ = 2l the area becomes A/2.
R′ = ρ(2l)/(A/2) = 4ρl/A = 4R. The resistance becomes four times.
In simple words: Doubling the length by stretching quadruples the resistance.

Teacher's Note:
1. Volume is constant when a wire is stretched.
2. Length doubled and area halved: the effect multiplies.

Question 16. [3 marks] (a) State Joule’s law of heating. (b) Why do we need a 5 A fuse for an electric iron which consumes 1 kW power at 220 V? (c) Why is it impracticable to connect an electric bulb and an electric heater in series?
Answer:
(a) The heat produced in a resistor is directly proportional to the square of the current, to the resistance and to the time: H = I²Rt.
(b) I = P/V = 1000/220 = 4.54 A, so a fuse slightly above this value, i.e. 5 A, is needed. It blows when the current exceeds the safe limit.
(c) In series the same current flows through both, but a bulb and a heater need widely different currents to work properly.
In simple words: A fuse rating is chosen just above the normal working current.

Teacher's Note:
1. Never choose a fuse much larger than the working current.
2. Series means the same current everywhere.

Question 17. [3 marks] (a) Derive the relation for the equivalent resistance when three resistors R₁, R₂ and R₃ are connected in parallel. (b) Find the minimum resistance that can be made using four resistors, each of 20 Ω.
Answer:
(a) The total current is I = I₁ + I₂ + I₃. The potential difference V is the same across each, so I = V/R_p, I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃.
Hence V/R_p = V/R₁ + V/R₂ + V/R₃, i.e. 1/R_p = 1/R₁ + 1/R₂ + 1/R₃.
(b) The minimum resistance is obtained by connecting all four in parallel: 1/R = 4/20, so R = 5 Ω.
In simple words: Joining all resistors in parallel gives the least resistance.

Teacher's Note:
1. The derivation uses current addition and equal voltage.
2. Minimum resistance means parallel; maximum means series.

Question 18. [3 marks] (a) What is the total resistance of n resistors each of resistance R connected in (i) series, (ii) parallel? (b) Calculate the resultant resistance of 3 resistors 3 Ω, 4 Ω and 12 Ω connected in parallel.
Answer:
(a) (i) R_s = R + R + … (n times) = nR.
(ii) 1/R_p = n/R, so R_p = R/n.
(b) 1/R_p = 1/3 + 1/4 + 1/12 = (4 + 3 + 1)/12 = 8/12
R_p = 12/8 = 1.5 Ω
In simple words: Equal resistors: series gives nR, parallel gives R/n.

Teacher's Note:
1. Use a common denominator of 12.
2. Parallel answer must be smaller than 3 Ω.

Question 19. [3 marks] Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (a) 13.5 Ω (b) 6 Ω?
Answer:
(a) 13.5 Ω: join two resistors in parallel, 1/R_p = 1/9 + 1/9 = 2/9, so R_p = 4.5 Ω; connect the third 9 Ω resistor in series with it: 9 + 4.5 = 13.5 Ω.
(b) 6 Ω: join two resistors in series, R_s = 18 Ω; connect the third 9 Ω in parallel with it: 1/R = 1/9 + 1/18 = 3/18, so R = 6 Ω.
In simple words: Mix series and parallel to reach 13.5 Ω and 6 Ω.

Teacher's Note:
1. Larger than 9 Ω needs a series part; smaller than 9 Ω needs a parallel part.
2. Check each result by recalculating.

Question 20. [5 marks] (A) What is meant by the series combination and parallel combination of resistances? (B) In the circuit diagram given, five resistances of 5 Ω, 20 Ω, 15 Ω, 20 Ω and 10 Ω are connected as shown to a 6 V battery. Calculate the total resistance in the circuit.

Answer:
(A) Series: resistors are joined end to end, so the same current flows through each, and R_s = R₁ + R₂ + R₃. Parallel: resistors are joined between the same two points, so each has the same potential difference, and 1/R_p = 1/R₁ + 1/R₂ + 1/R₃.
(B) The 5 Ω and 20 Ω resistors are in parallel: 1/R_A = 1/5 + 1/20 = 5/20, so R_A = 4 Ω.
The 15 Ω, 20 Ω and 10 Ω resistors are in parallel: 1/R_B = 1/15 + 1/20 + 1/10 = (4 + 3 + 6)/60 = 13/60, so R_B = 60/13 Ω.
The two groups are in series: R = 4 + 60/13 = 8.6 Ω.
In simple words: Solve each parallel group first, then add the two groups.

Teacher's Note:
1. Identify which resistors share the same two junctions.
2. Series groups add; parallel groups use reciprocals.

Question 21. [3 marks] Draw a schematic diagram of a circuit consisting of a battery of four dry cells of 1.5 V each, a 2 Ω resistor, a 6 Ω resistor, 16 Ω resistor and a plug key all connected in series. Put an ammeter to measure the current in the circuit and a voltmeter across the 16 Ω resistor to measure potential difference across its two ends. Use Ohm’s law to determine: (A) ammeter reading, and (B) voltmeter reading when key is closed.

Answer:
The circuit is drawn alongside: battery, key, ammeter and the three resistors in series, with the voltmeter across the 16 Ω resistor.
Battery voltage V = 4 × 1.5 = 6 V; total resistance R = 2 + 6 + 16 = 24 Ω.
(A) Ammeter reading I = V/R = 6/24 = 0.25 A.
(B) Voltmeter reading = IR = 0.25 × 16 = 4 V.
In simple words: The same 0.25 A flows through every resistor; the voltmeter reads the drop across 16 Ω.

Teacher's Note:
1. Cells in series add their voltages.
2. Voltmeter is in parallel with the 16 Ω resistor.

Question 22. [5 marks] Draw a circuit diagram for the experimental verification of Ohm’s law and state the function of each component used.

Answer:
The circuit (see figure) has a battery (cell), a plug key, an ammeter in series, the resistor (nichrome wire) and a rheostat in series, and a voltmeter in parallel with the resistor.
Battery: supplies the potential difference. Key: opens or closes the circuit. Ammeter: measures the current through the resistor (in series). Voltmeter: measures the potential difference across the resistor (in parallel). Rheostat: changes the current (and voltage) in steps. Resistor: the conductor whose resistance is found.
Plotting V against I gives a straight line through the origin, which verifies Ohm's law.
In simple words: The ammeter is in series; the voltmeter is across the resistor.

Teacher's Note:
1. Draw correct symbols and polarity.
2. Note the zero errors of the meters.

Question 23. [5 marks] (i) Write an expression for the resistivity of a substance. (ii) State the SI unit of resistivity. (iii) Distinguish between resistance and resistivity. (iv) Name two factors on which the resistivity of a substance depends and two factors on which it does not depend.
Answer:
(i) ρ = RA/l.
(ii) The SI unit of resistivity is ohm-metre (Ω m).
(iii) Resistivity is a characteristic property of the material and does not depend on its dimensions, whereas resistance depends on the dimensions (length and area) of the conductor.
(iv) Resistivity depends on (a) the nature of the material and (b) the temperature. It does not depend on (a) the length and (b) the area of cross-section of the conductor.
In simple words: Resistivity belongs to the material; resistance belongs to the particular wire.

Teacher's Note:
1. Link ρ with R = ρl/A.
2. Keep the two lists clearly separate.

Question 24. [5 marks] Explain the following: (i) Why is tungsten used almost exclusively for filament of electric lamps? (ii) Why are the conductors of electric heating devices, such as bread-toaster and electric iron, made of an alloy rather than a pure metal? (iii) Why is series arrangement not used for domestic circuits? (iv) How does the resistance of a wire vary with its area of cross-section? (v) Why are copper and aluminium wires usually used for the transmission of electric current?
Answer:
(i) Tungsten has a high resistivity and a very high melting point, so it does not melt at the high temperature at which it glows.
(ii) Alloys have a higher resistivity than pure metals and a higher melting point, and they do not oxidise (burn) readily at high temperature.
(iii) In series the resistance increases and the current decreases, and if one component fails the other components also stop working.
(iv) The resistance of a wire is inversely proportional to its area of cross-section.
(v) Copper and aluminium have low resistivity, so they conduct well with little loss of energy.
In simple words: Each answer links a property of the material to its use.

Teacher's Note:
1. Give one clear property for each use.
2. These are standard five-mark explanations.

Question 25. [5 marks] (a) Define electric power. Express it in terms of potential difference, V and resistance, R. (b) An electrical fuse is rated at 2 A. What is meant by this statement? (c) An electric iron of 1 kW is operated at 220 V. Find which of the fuse that are respectively rated at 1 A, 3 A and 5 A can be used for it.
Answer:
(a) Electric power is the rate at which electrical energy is consumed (or work is done): P = W/t = VI. Using V = IR, P = V²/R.
(b) It means the maximum current that can flow through the fuse is 2 A; if the current exceeds 2 A the fuse wire melts.
(c) I = P/V = 1000/220 = 4.54 A. The 1 A and 3 A fuses would blow, so the 5 A fuse should be used.
In simple words: Fuse rating must be just above the working current.

Teacher's Note:
1. Calculate the current before choosing.
2. A 1 A or 3 A fuse would blow at 4.54 A.