CBSE Class 10 Science Electricity VBQs Set 02

Question. Two resistors connected to 100 V supply in parallel draw 10 A current from the supply. If the power dissipation in one resistor is 600 W, find
(i) power dissipation in the other.
(ii) resistance of each. 
Answer:
(i) Total power \( P = V \times I = 100 \text{ V} \times 10 \text{ A} = 1000 \text{ W} \).
Power dissipation in the first resistor, \( P_1 = 600 \text{ W} \).
Power dissipation in the second resistor, \( P_2 = P - P_1 = 1000 \text{ W} - 600 \text{ W} = 400 \text{ W} \).
(ii) Resistance of first resistor, \( R_1 = \frac{V^2}{P_1} = \frac{100^2}{600} = \frac{10000}{600} = \frac{50}{3} \approx 16.67 \text{ } \Omega \).
Resistance of second resistor, \( R_2 = \frac{V^2}{P_2} = \frac{100^2}{400} = \frac{10000}{400} = 25 \text{ } \Omega \).

Question. Three resistors \( 3 \text{ } \Omega \), \( 6 \text{ } \Omega \) and \( 9 \text{ } \Omega \) are connected to a battery. In which of them power dissipation be maximum, if, they are all connected in (i) parallel (ii) series. Give reasons. 
Answer:
(i) In parallel, potential difference \( V \) is same. Since \( P = \frac{V^2}{R} \), \( P \propto \frac{1}{R} \). Therefore, power dissipation is maximum in the \( 3 \text{ } \Omega \) resistor.
(ii) In series, current \( I \) is same. Since \( P = I^2 R \), \( P \propto R \). Therefore, power dissipation is maximum in the \( 9 \text{ } \Omega \) resistor.

Question. List in a tabular form three differences between a voltmeter and an ammeter.
Answer:

• Ammeter: Used to measure electric current; Connected in series; Has very low resistance.
• Voltmeter: Used to measure potential difference; Connected in parallel; Has very high resistance.

Question. Name the physical quantity which is (i) same (ii) different in all the bulbs when three bulbs of :
(a) same wattage are connected in series.
(b) same wattage are connected in parallel.
(c) different wattage are connected in series.
(d) different wattage are connected in parallel. 
Answer:
(a) Same: Current; Different: Potential Difference (though if wattage is same, voltage will also be same).
(b) Same: Potential Difference; Different: Current (though if wattage is same, current will also be same).
(c) Same: Current; Different: Potential Difference.
(d) Same: Potential Difference; Different: Current.

Question. Two resistors with resistances \( 5 \text{ } \Omega \) and \( 10 \text{ } \Omega \) are to be connected to a battery of emf 6 V so as to obtain:
(i) minimum current (ii) maximum current
(a) How will you connect the resistances in each case ?
(b) Calculate the strength of the total current in the circuit in the two cases. 
Answer:
(a) (i) For minimum current, connect resistors in series. (ii) For maximum current, connect resistors in parallel.
(b) Case (i) Series: \( R_s = 5 + 10 = 15 \text{ } \Omega \).
\( \implies I_{min} = \frac{6}{15} = 0.4 \text{ A} \).
Case (ii) Parallel: \( R_p = \frac{5 \times 10}{5 + 10} = \frac{50}{15} = \frac{10}{3} \text{ } \Omega \).
\( \implies I_{max} = \frac{6}{10/3} = \frac{18}{10} = 1.8 \text{ A} \).

Question. (a) Write any three difference between the series and parallel combination of resistance.
(b) A set of ‘n’ identical resistors each resistance R are connected in series and the effective resistance is found to be ‘X’. When these are connected in parallel, the effective resistance is found to be ‘Y’. Find the ratio of X and Y. 
Answer:
(a) In series: \( R_{eq} = \sum R \), current is same, voltage divides. In parallel: \( 1/R_{eq} = \sum 1/R \), voltage is same, current divides.
(b) Series: \( X = nR \).
Parallel: \( Y = \frac{R}{n} \).
Ratio \( \frac{X}{Y} = \frac{nR}{R/n} = n^2 \).

Question. (a) Though same current flows through the electric line wires and the filament of bulb, yet only the filament glows. Why?
(b) The temperature of the filament of bulb is 2700 °C when it glows. Why does it not get burnt up at such high temperature?
(c) The filament of an electric lamp, which draws a current of 0.25 A is used for four hours. Calculate the amount of charge flowing through the circuit.
(d) An electric iron is rated 2 kW at 220 V. Calculate the capacity of the fuse that should be used for the electric iron. 
Answer:
(a) The filament has very high resistance compared to line wires. Heat produced \( H = I^2Rt \). Due to high R, filament produces enough heat to become white hot and glow.
(b) Filament is made of Tungsten which has a very high melting point (approx 3380 °C) and the bulb is filled with inactive gases like Nitrogen/Argon to prevent oxidation.
(c) \( Q = I \times t = 0.25 \text{ A} \times (4 \times 3600 \text{ s}) = 0.25 \times 14400 = 3600 \text{ C} \).
(d) \( I = \frac{P}{V} = \frac{2000}{220} \approx 9.09 \text{ A} \). So, a 10 A fuse should be used.

Question. (a) An electric iron consumes energy at a rate of 880 W when heating is at the maximum rate and 330 W when the heating is at the minimum. If the source voltage is 220 V, calculate the current and resistance in each case.
(b) What is heating effect of electric current?
(c) Find an expression for the amount of heat produced when a current passes through a resistor for some time.
Answer:
(a) Maximum: \( I = \frac{880}{220} = 4 \text{ A} \), \( R = \frac{220}{4} = 55 \text{ } \Omega \).
Minimum: \( I = \frac{330}{220} = 1.5 \text{ A} \), \( R = \frac{220}{1.5} \approx 146.67 \text{ } \Omega \).
(b) The production of heat in a conductor due to the flow of electric current through it is called the heating effect of current.
(c) From \( V = \frac{W}{Q} \), \( W = VQ = VIt \). Since \( V = IR \), \( H = (IR)It = I^2Rt \).

ASSERTION AND REASON QUESTIONS

In the following Questions, the Assertion and Reason have been put forward. Read the statements carefully and choose the correct alternative from the following:
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.

Question. Assertion: The metals and alloys are good conductors of electricity.
Reason: Bronze is an alloy of copper and tin and it is not a good conductor of electricity. 
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (c) Assertion is true but the Reason is false.

Question. Assertion: Alloys are commonly used in electrical heating devices like electric iron and heater.
Reason: Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points than their constituent metals.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (c) Assertion is true but the Reason is false.

Question. Assertion: At high temperatures, metals wires have a greater chance of short circuiting.
Reason: Both resistance and resistivity of a material vary with temperature.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.

Question. Assertion: Conductors allow the current to flow through themselves.
Reason: They have free charge carriers.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

Question. Assertion: In an open circuit, the current passes from one terminal of the electric cell to another.
Reason: Generally, the metal disc of a cell acts as a positive terminal.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (d) The statement of the Assertion is false but the Reason is true.

Question. Assertion: The statement of Ohm’s law is V = IR
Reason: V = IR is the equation which defines resistance.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (c) Assertion is true but the Reason is false.

Question. Assertion: Bending of wire decrease the resistance of electric wire.
Reason: The resistance of a conductor depends on length, thickness, nature of material and temperature of the conductor.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (d) The statement of the Assertion is false but the Reason is true.

Question. Assertion: If a graph is plotted between potential difference and current a linear graph is obtained.
Reason: Current is directly proportional to the potential difference.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

Question. Assertion: A cell converts chemical energy into electrical energy.
Reason: A cell maintains a potential difference across its terminals due to chemical reactions. 
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

Question. Assertion: Thicker wires have smaller resistance and the thinner wires have greater resistance.
Reason: Resistance is inversely proportional to the area of cross-section of wire.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

Question. What does an electric circuit mean?
Answer: Electric circuit is a continuous and closed path made of conducting wires, through which the electric current flows. It comprises a cell, ammeter, voltmeter, plug key, etc.

Question. Define the unit of current.
Answer: SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross-section of wire for 1 second, then it would be equal to 1 ampere.
So, 1 ampere = \( \frac{1 \text{ C}}{1 \text{ s}} \) i.e. \( 1\text{A} = 1 \text{ Cs}^{-1} \)

Question. Calculate the number of electrons constituting one coulomb of charge.
Answer: Given \( q = 1 \text{ C} \), \( e = 1.6 \times 10^{-19} \text{ C} \), \( n = ? \), \( q = ne \)
\( 1 \text{ C} = n \times 1.6 \times 10^{-19} \text{ C} \)
\( n = \frac{1}{(1.6 \times 10^{-19})} = 6.25 \times 10^{18} \text{ electrons} \)

Question. Name a device that helps to maintain a potential difference across a conductor. 
Answer: Cell or battery eliminator.

Question. What is meant by saying that the potential difference between two points is 1 V?
Answer: As we know that \( V = \frac{W}{q} \)
Thus, the potential difference between two points is one volt when one joule of work is done to carry a charge of one coulomb between the two points in the electric field.

Question. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer: \( q = 1 \text{ C} \); \( V = 6 \text{ V} \),
\( V = \frac{W}{q} \)
\( \implies W = q \times V \)
So, Work done = \( 1 \text{ C} \times 6 \text{ V} = 6 \text{ J} \)
As, Energy = Work done
\( \implies \text{Energy} = 6 \text{ J} \)

Question. On what factors does the resistance of a conductor depend? 
Answer: Resistance of a conductor depends upon:
(i) Resistivity of the material.
(ii) Length of the conductor.
(iii) Cross-sectional area of the conductor.
(iv) Temperature of the conductor.

Question. Will current flow more easily through a thick wire or thin wire of the same material, when connected to the same source? Why?
Answer: The current flows more easily through a thick wire than through a thin wire because the resistance of thick wire is less than that of a thin wire as \( R \propto 1/A \).

Question. Let 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: As we know that \( I = \frac{V}{R} \)
if \( V' = \frac{V}{2} \)
\( \implies I' = \frac{V'}{R} = \frac{V}{2R} = \frac{I}{2} \)
Hence, the current through it also becomes half of its previous value.

Question. Why are the coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer: The coils of electric toaster and electric iron are made of an alloy rather than a pure metal because of the following reasons:
(i) The resistivity of an alloy is higher than that of a pure metal.
(ii) It has high melting point and does not oxidise.

Question. Use the data in Table 11.2 of NCERT book to answer the following:
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer: (a) Iron because its resistivity is less than mercury.
(b) Silver is the best conductor as it has the least resistivity.

Question. Judge the equivalent resistance when the following are connected in parallel.
(a) \( 1 \text{ } \Omega \) and \( 10^6 \text{ } \Omega \)
(b) \( 1 \text{ } \Omega \) and, \( 10^3 \text{ } \Omega \), and, \( 10^6 \text{ } \Omega \).
Answer: Equivalent resistance in parallel combination of resistors is always less than the least resistance of any resistor in the circuit. Hence, in both the given cases, the equivalent resistance is less than \( 1 \text{ } \Omega \).

Question. An electric lamp of \( 100 \text{ } \Omega \), a toaster of resistance \( 50 \text{ } \Omega \), and a water filter of resistance \( 500 \text{ } \Omega \) 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 flows through it?
Answer: \( R_1 = 100 \text{ } \Omega, R_2 = 50 \text{ } \Omega, R_3 = 500 \text{ } \Omega \)
All the applications are connected in parallel, so
\( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)
\( = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} \)
\( = \frac{5 + 10 + 1}{500} = \frac{16}{500} \)
\( R = \frac{500}{16} = \frac{125}{4} \text{ } \Omega \)
Current through all the appliances
\( I = \frac{V}{R} = \frac{220}{125/4} = \frac{220 \times 4}{125} = 7.04 \text{ A} \)
Now if only electric iron is connected to the same source such that it takes as much current as all three appliances, i.e. \( I = 7.04 \text{ A} \), its resistance should be equal to \( \frac{125}{4} \text{ } \Omega \), i.e. \( 31.25 \text{ } \Omega \).

Question. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer: Advantages of connecting electrical devices in parallel:
(i) When the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that of battery which is not possible in series connection.
(ii) Each appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, same current flows through all devices, irrespective of their resistances.
(iii) If one appliance fails to work, other will continue to work properly. If they are connected in parallel.

Question. What is (a) the highest (b) the lowest total resistance that can be secured by combination of four coils of resistances \( 4 \text{ } \Omega, 8 \text{ } \Omega, 12 \text{ } \Omega, 24 \text{ } \Omega \)?
Answer: (a) The highest resistance is secured by combining all four coils of resistance in series.
\( R_s = 4 \text{ } \Omega + 8 \text{ } \Omega + 12 \text{ } \Omega + 24 \text{ } \Omega = 48 \text{ } \Omega \)
(b) The lowest resistance is secured by combining all four coils of resistance in parallel.
\( \frac{1}{R_p} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} \)
\( R_p = 2 \text{ } \Omega \)

Question. Why does the cord of an electric heater not glow while the heating element does?
Answer: The cord of an electric heater is made up of metallic wire such as copper or aluminium which has low resistance while the heating element is made up of an alloy which has more resistance than its constituent metals. Also heat produced 'H' is \( H = I^2Rt \).
Thus, for the same current \( H \propto R \), more heat is produced by heating element as it has more resistance, and it glows.

Question. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer: Given \( q = 96000 \text{ C} \), \( V = 50 \text{ V} \), \( t = 1 \text{ h} \)
\( H = I^2Rt = VIt = Vq = 50 \times 96000 = 48 \times 10^5 \text{ J} \)

Question. An electric iron of resistance \( 20 \text{ } \Omega \) takes a current of 5 A. Calculate the heat developed in 30 s.
Answer: Given \( R = 20 \text{ } \Omega, I = 5 \text{ A}, t = 30 \text{ s} \)
\( H = I^2Rt = (5)^2 \times 20 \times 30 = 15000 \text{ J} = 1.5 \times 10^4 \text{ J} \)

Question. What determines the rate at which energy is delivered by a current?
Answer: Electric power determines the rate at which energy is delivered by a current.

Question. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer: Given \( I = 5 \text{ A}, V = 220 \text{ V}, t = 2 \text{ h} \)
Power, \( P = VI = 220 \times 5 = 1100 \text{ W} \)
Energy consumed, \( E = P \times t = 1100 \times 2 \times 60 \times 60 = 1100 \times 7200 = 7.92 \times 10^6 \text{ J} \)

EXERCISES

Question. A 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

Question. Which of the following terms does not represent electrical power in a circuit?
(a) \( I^2R \)
(b) \( IR^2 \)
(c) \( VI \)
(d) \( V^2/R \)
Answer: (b) \( IR^2 \)

Question. An 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

Question. Two conducting wires of the same material and of equal lengths and 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