CBSE Class 10 Science Electricity Worksheet

Question. A letter A is constructed of a uniform wire of resistance 1 ohm per cm. The sides of the letter are 20 cm and the cross piece in the middle is 10 cm long. The resistance between the ends of the legs will be
(a) 32.4 ohm
(b) 28.7 ohm
(c) 26.7 ohm
(d) 24.7 ohm
Answer : C

Question. Three resistances of 2, 3 and 5 W are connected in parallel to a 10 V battery of negligible internal resistance. The potential difference across the 3 W resistance will be
(a) 2 V
(b) 3 V
(c) 5 V
(d) 10 V
Answer : D

Question. You are given n identical wires, each of resistance R. When these are connected in parallel, the equivalent resistance is X. When these will be connected in series, then the equivalent resistance will be
(a) X/n²
(b) n²X
(c) X/n
(d) nX
Answer : B

Question. 20 coulomb charge is flowing in 0.5 second from a point in an electric circuit then value of electric current in amperes will be
(a) 10
(b) 40
(c) 0.005
(d) 0.05
Answer : B

Question. A cylindrical rod is reformed to twice its length with no change in its volume. If the resistance of the rod was R, the new resistance will be
(a) R
(b) 2R
(c) 4R
(d) 8R
Answer : C

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 isRl, then the ratio R/Rl is
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Answer : D

Question. 2 ampere current is flowing through a conductor from a 10 volt emf source then resistance of conductor is
(a) 20 W
(b) 5 W
(c) 12 W
(d) 8 W
Answer : B

Question. Three resistors of 4.0 W , 6.0 W and 10.0 W are connected in series. What is their equivalent resistance
(a) 20 W
(b) 7.3 W
(c) 6.0 W
(d) 4.0 W
Answer : C

Question. Two unequal resistances are connected in parallel. Which of the following statement is true
(a) current in same in both
(b) current is larger in higher resistance
(c) voltage-drop is same across both
(d) voltage-drop is lower in lower resistance
Answer : C

Question. A wire of resistance R is cut into ten equal parts which are then joined in parallel. The new resistance is
(a) 0.01 R
(b) 0.1 R
(c) 10 R
(d) 100 R
Answer : A

Question. A 24 V potential difference is applied is applied across a parallel combination of four 6 ohm resistor. The current in each resistor is
(a) 1 A
(b) 4 A
(c) 16 A
(d) 36 A
Answer : B

Question. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R₁, R₂ and R₃ respectively (Figure).
Which of the following is true?
(a) R₁ > R₂ = R₃ 
(b) R₁ > R₂ > R₃
(c) R₃ > R₂ = R₁
(d) R₂ > R₃ > R₁
Answer : C

Question. A current of 4.8 A is flowing in a conductor. The number of electrons passing per second through the conductor will be
(a) 3 x 10²⁰
(b) 76.8 x 10²⁰
(c) 7.68 x 10-¹⁹
(d) 3 x 10¹⁹
Answer : D

Question. How much work is done in moving a charge of 2 C from a point of 118 V to a point of 128 V?
(a) 20 J
(b) 30 J
(c) 40 J
(d) 10 J
Answer : A

Question. Two bulbs have the following ratings:
1. 40 W, 220 V
2. 20 W, 100 V
The ratio of their resistance is
(a) 1: 2
(b) 2 :1
(c) 1:1
(d) 1: 3
Answer : B

Question. A current of 1 A is drawn by a filament of an electric bulb. Number or electrons passing through a cross section of the filament in 16 seconds would be roughly
(a) 10²⁰
(b) 10¹⁶
(c) 10¹⁸
(d) 10²³
Answer : A

Question. If a wire of resistance R is melted and recast to half of its length, the new resistance of the wire will be
(a) R/4
(b) R/2
(c) R
(d) 2R
Answer : A

Question. A circular conductor is made of a uniform wire of resistance 2x10^-3 ohm/metre and the diameter of this circular conductor is 2 metres. Then the resistance measured between the ends of the diameter is (in ohms)
(a) p x 10^-3
(b) 2p x 10^-3
(c) 4p x 10^-3
(d) 4x 10^-3
Answer : C

Question. The length of a wire is doubled. By what factor does the resistance change
(a) 4 time as large
(b) twice as large
(c) unchanged
(d) half as large
Answer : D

Question. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A
Answer : D

Question. What is the resultant resistance when number of resistors are connected in parallel? 

Answer :  A circuit in which two or more resistors are connected across common points so as to provide separate paths is called parallel circuit.

In this case, the same potential difference will be maintained between the two ends of every resistor and the current will divide itself in various branches.

Let the resistors R₁, R₂ and R₃ be joined in parallel to the points A and B. Let the current I reaching A divide itself into three parts I₁, I₂ and I₃ along R₁, R₂ and R₃ respectively. Let V be the potential difference between the points A and B. The current flowing in the individual resistors are then given by :

Question. V-I graph for two wires A and B are shown in the figure. If both wires are of same length and same thickness, which of the two is made of a material of high resistivity? Give justification for your answer.

Answer :  Greater than slope of V-I graph, greater will be the resistance of given metallic wire. In the given graph, wire A has greater slope then B. Hence, wire A has greater resistance. For the wires of same length and same thickness, resistance depends on the nature of material of the wire, i.e.

Question. A wire of resistance 20 Ω is bent to form a closed square. What is the resistance across a diagonal of the square?
Answer :  

Question. Calculate the number of electrons constituting one coulomb of charge.
Answer :  One electron possesses a charge of 1.6 ×10^-19C, i.e., 1.6 ×10^-19C of charge is contained in 1 electron.
∴ 1 C of charge is contained in 1/1.6 x 10^-19 = 6.25 x 10¹⁸ = 6 x 10¹⁸
Therefore, 6 x 10¹⁸ electrons constitute one coulomb of charge. 

Question.  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 : Fuse wire is a safety device connected in series with the live wire of the circuit, since it has high  resistivity and low melting point. It melts when a sudden urge of large current passes through it and disconnects the entire circuit from the electrical supply. But in case if we use a larger rating fuse wire instead of a defined rating fuse wire then it will not protect the circuit as high current will easily pass through it and it will not melt.

Question.  Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer : Energy consumed by 250 W TV in 1 h = 250 × 1 = 250 W h
Energy consumed by 1200 W toaster in 10 minutes = 1200 x 10/60 = 200 W h
Here total energy consumed by the TV is more than toaster

Question.  Why does the cord of an electric heater not glow while the heating element does? 
Answer : The cord of the electric heater is made up of copper or aluminium which has lower resistance and higher conductance, therefore very less heat energy is produced.
Heating element is made up of alloys which have high resistance and lot of heat is produced, and only a little amount of electric energy makes it to glow

Long Answers :

or I μ V or V μ I or V = RI

where R is a constant called resistance. Resistance is the property of a body to oppose the flow of current. R the resistance depends upon the nature of the conductor, its temperature and its dimensions (length, area)

R = V/I or I = V/Ri.e. I α V and I α I/R

 Experimental Verification of Ohm's Law: To verify Ohm's law, take a resistor R, connect voltmeter across it. Connect an ammeter, battery, key and rheostat to it as shown in Fig. Put in the key K. Read the value of potential difference across resistor R with the help of voltmeter and the current flowing through resistor with the help of ammeter. Note the readings. Vary the current in the circuit by sliding rheostat and go on noting reading in voltmeter and ammeter. Plot a graph between V and I on graph paper. It will come out to be straight line as shown in fig.

Question. A. Two identical wires one of nichrome and other of copper are connected in series and a current (I) is passed through them. State the change observed in the temperatures of the two wires. Justify your answer. State the law which explains the above observation.
B. An electric bulb is rated at 60 W, 240 V. Calculate its resistance. If the voltage drops to 192 V, calculate the power consumed and the current drawn by the bulb. (Assume that the resistance of the bulb remains unchanged.)
Answer :  A. The resistivity of nichrome is more than that of copper so its resistance is also high. Therefore, large amount of heat is produced in the nichrome wire for the same current as compared to that of copper wire. Accordingly, more change in temperature is observed in the nichrome wire. This is explained by Joule’s law of heating. Joule’s law of heating: It states that the amount of heat produced in a conductor is

Question. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer : The energy given to each coulomb of charge is equal to the amount of work which is done in moving it.
Now we know that,
Potential difference = Work Done/Charge
∴ Work done = Potential difference × charge
Where, Charge = 1 C and Potential difference = 6 V
∴ Work done = 6×1
= 6 Joule. 

Question.  On what factors does the resistance of a conductor depend?
Answer : The resistance of a conductor depends upon the following factors:
→ Length of the conductor
→ Cross-sectional area of the conductor
→ Material of the conductor
→ Temperature of the conductor 

Page No: 200

Question. What does an electric circuit mean?
Answer: A continuous and closed path of an electric current is called an electric circuit. An electric circuit consists of electric devices, source of electricity and wires that are connected with the help of a switch.

Question. Define the unit of current.
Answer: The unit of electric current is ampere (A). 1 A is defined as the flow of 1 C of charge through a wire in 1 s.

Question. Calculate the number of electrons constituting one coulomb of charge.
Answer: One electron possesses a charge of \( 1.6 \times 10^{-19} \text{ C} \), i.e., \( 1.6 \times 10^{-19} \text{ C} \) of charge is contained in 1 electron.
\( \therefore 1 \text{ C} \) of charge is contained in \( \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} \approx 6 \times 10^{18} \)
Therefore, \( 6 \times 10^{18} \) electrons constitute one coulomb of charge.

Page No: 202

Question. Name a device that helps to maintain a potential difference across a conductor.
Answer: Any source of electricity like battery, cell, power supply, etc. helps to maintain a potential difference across a conductor.

Question. What is meant by saying that the potential difference between two points is 1 V?
Answer: If 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.

Question. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer: The energy given to each coulomb of charge is equal to the amount of work which is done in moving it.
Now we know that,
Potential difference = Work Done/Charge
\( \dots \) Work done = Potential difference \( \times \) charge
Where, Charge = 1 C and Potential difference = 6 V
\( \dots \) Work done = \( 6 \times 1 = 6 \text{ Joule} \).

Page No: 209

Question. On what factors does the resistance of a conductor depend?
Answer: The resistance of a conductor depends upon the following factors:
→ Length of the conductor
→ Cross-sectional area of the conductor
→ Material of the conductor
→ Temperature of the conductor

Question. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer: The current will flow more easily through thick wire. It is because the resistance of a conductor is inversely proportional to its area of cross - section. If thicker the wire, less is resistance and hence more easily the current flows.

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: According to Ohm’s law
\( V = IR \)
\( \Rightarrow I = V/R \dots (1) \)
Now Potential difference is decreased to half
\( \therefore \) New potential difference \( V' = V/2 \)
Resistance remains constant
So the new current \( I' = V'/R \)
\( = (V/2)/R \)
\( = (1/2) (V/R) \)
\( = (1/2) I = I/2 \)
Therefore, the amount of current flowing through the electrical component is reduced by half.

Question. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer: The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.

Question. Use the data in Table 12.2 to answer the following -
Table 12.2 Electrical resistivity of some substances at 20°C
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

Answer:
(a) Resistivity of iron = \( 10.0 \times 10^{-8} \ \Omega \)
Resistivity of mercury = \( 94.0 \times 10^{-8} \ \Omega \)
Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.
(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.

Page No: 213

Question. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 \( \Omega \) resistor, an 8 \( \Omega \) resistor, and a 12 \( \Omega \) resistor, and a plug key, all connected in series.
Answer: Three cells of potential 2 V, each connected in series therefore the potential difference of the battery will be 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors of resistances 5 \( \Omega \), 8 \( \Omega \) and 12 \( \Omega \) respectively connected in series and a battery of potential 6 V and a plug key which is closed means the current is flowing in the circuit.

Question. Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure potential difference across the 12 \( \Omega \) resistor. What would be the readings in the ammeter and the voltmeter?
Answer: An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.

The resistances are connected in series.
Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,
\( V = IR \)
Where,
Potential difference, \( V = 6 \text{ V} \)
Current flowing through the circuit/resistors = \( I \)
Resistance of the circuit, \( R = 5 + 8 + 12 = 25 \, \Omega \)
\( I = V/R = 6/25 = 0.24 \text{ A} \)
Potential difference across \( 12 \, \Omega \) resistor = \( V_1 \)
Current flowing through the \( 12 \, \Omega \) resistor, \( I = 0.24 \text{ A} \)
Therefore, using Ohm’s law, we obtain
\( V_1 = IR = 0.24 \times 12 = 2.88 \text{ V} \)
Therefore, the reading of the ammeter will be \( 0.24 \text{ A} \).
The reading of the voltmeter will be \( 2.88 \text{ V} \).

Page No: 216

Question. Judge the equivalent resistance when the following are connected in parallel − (a) 1 \( \Omega \) and \( 10^6 \ \Omega \), (b) 1 \( \Omega \) and \( 10^3 \ \Omega \) and \( 10^6 \ \Omega \).
Answer: (a) When 1 \( \Omega \) and \( 10^6 \ \Omega \) are connected in parallel:
Let \( R \) be the equivalent resistance.
\[ \therefore \frac{1}{R} = \frac{1}{1} + \frac{1}{10^6} \]
\[ R = \frac{10^6}{1 + 10^6} \approx \frac{10^6}{10^6} = 1 \, \Omega \]
Therefore, equivalent resistance \( \approx 1 \ \Omega \)

(b) When 1 \( \Omega \), \( 10^3 \ \Omega \) and \( 10^6 \ \Omega \) are connected in parallel:
Let \( R \) be the equivalent resistance.
\[ \frac{1}{R} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6} = \frac{10^6 + 10^3 + 1}{10^6} \]
\[ R = \frac{1000000}{1001001} \approx 0.999 \ \Omega \]
Therefore, equivalent resistance \( \approx 0.999 \ \Omega \).

Question. An electric lamp of 100 \( \Omega \), a toaster of resistance 50 \( \Omega \), and a water filter of resistance 500 \( \Omega \) 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 through it?
Answer: Resistance of electric lamp, \( R_1 = 100 \ \Omega \)
Resistance of toaster, \( R_2 = 50 \ \Omega \)
Resistance of water filter, \( R_3 = 500 \ \Omega \)
Potential difference of the source, \( V = 220 \text{ V} \)
These are connected in parallel, as shown in the following figure.

Let \( R \) be the equivalent resistance of the circuit.
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} \]
According to Ohm's law,
\( V = IR \)
\( I = \frac{V}{R} \)
Where,
Current flowing through the circuit = \( I \)
\[ I = \frac{220}{\frac{500}{16}} = \frac{220 \times 16}{500} = 7.04 \text{ A} \]
\( 7.04 \text{ A} \) of current is drawn by all the three given appliances.
Therefore, current drawn by an electric iron connected to the same source of potential \( 220 \text{ V} = 7.04 \text{ A} \)
Let \( R' \) be the resistance of the electric iron. According to Ohm's law,
\( V = IR' \)
\[ R' = \frac{V}{I} = \frac{220}{7.04} = 31.25 \ \Omega \]
Therefore, the resistance of the electric iron is \( 31.25 \ \Omega \) and the current flowing through it is \( 7.04 \text{ A} \).

Question. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer: There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.
The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

Question. How can three resistors of resistances 2 \( \Omega \), 3 \( \Omega \) and 6 \( \Omega \) be connected to give a total resistance of (a) 4 \( \Omega \), (b) 1 \( \Omega \)?
Answer: There are three resistors of resistances 2 \( \Omega \), 3 \( \Omega \), and 6 \( \Omega \) respectively.
(a) The following circuit diagram shows the connection of the three resistors.

Here, 6 \( \Omega \) and 3 \( \Omega \) resistors are connected in parallel.
Therefore, their equivalent resistance will be given by
\[ R_p = \frac{6 \times 3}{6 + 3} = 2 \ \Omega \]
This equivalent resistor of resistance 2 \( \Omega \) is connected to a 2 \( \Omega \) resistor in series.
Therefore, the equivalent resistance of the circuit = \( 2 \ \Omega + 2 \ \Omega = 4 \ \Omega \)
Hence the total resistance of the circuit is 4 \( \Omega \).

(b) The following circuit diagram shows the connection of the three resistors.

All the resistors are connected in parallel. Therefore, their equivalent resistance will be given as
\[ \frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1 \ \Omega \]
Therefore, the total resistance of the circuit is 1 \( \Omega \).

Question. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 \( \Omega \), 8 \( \Omega \), 12 \( \Omega \), 24 \( \Omega \)?
Answer: There are four coils of resistances 4 \( \Omega \), 8 \( \Omega \), 12 \( \Omega \) and 24 \( \Omega \) respectively.
(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum \( 4 + 8 + 12 + 24 = 48 \ \Omega \)
(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by
\[ \frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{6+3+2+1}{24} = \frac{12}{24} \]
\( R = 2 \ \Omega \)
Therefore, 2 \( \Omega \) is the lowest total resistance.

Page No: 218

Question. Why does the cord of an electric heater not glow while the heating element does?
Answer: The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminium is very low so it does not glow.

Question. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer: Given Charge, Q = 96000C
Time, t= 1hr = 60 x 60= 3600s
Potential difference, V= 50volts
Now we know that H= VIt
So we have to calculate I first
As I= Q/t
\( \therefore \) I = 96000/3600 = 80/3 A
\[ H = 50 \times \frac{80}{3} \times 3600 = 4.8 \times 10^6 \text{ J} \]
Therefore, the heat generated is \( 4.8 \times 10^6 \text{ J} \).

Question. An electric iron of resistance 20 \( \Omega \) takes a current of 5 A. Calculate the heat developed in 30 s.
Answer: The amount of heat (H) produced is given by the joule's law of heating as H= VIt
Where,
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current x Resistance = 5 x 20 = 100 V
\( H = 100 \times 5 \times 30 = 1.5 \times 10^4 \text{ J} \).
Therefore, the amount of heat developed in the electric iron is \( 1.5 \times 10^4 \text{ J} \).

Page No: 220

Question. What determines the rate at which energy is delivered by a current?
Answer: The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

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: Power (P) is given by the expression, P = VI
Where,
Voltage, V = 220 V
Current, I = 5 A
P= 220 x 5 = 1100 W
Energy consumed by the motor = Pt
Where,
Time, t = 2 h = 2 x 60 x 60 = 7200 s
\( \therefore \) P = 1100 x 7200 = \( 7.92 \times 10^6 \text{ J} \)
Therefore, power of the motor = 1100 W
Energy consumed by the motor = \( 7.92 \times 10^6 \text{ J} \)

Page No: 221

Exercise

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 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: (c) 1:4

Question. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

Question. A copper wire has diameter 0.5 mm and resistivity of \( 1.6 \times 10^{-8} \ \Omega \text{ m} \). What will be the length of this wire to make its resistance 10 \( \Omega \)? How much does the resistance change if the diameter is doubled?
Answer: Area of cross-section of the wire, A =\( \pi \) (d/2) 2
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 \( \Omega \)
We know that
\( R = \rho \frac{l}{A} \)
\( l = \frac{RA}{\rho} \)
\( = \frac{10 \times 3.14 \times \left(\frac{0.0005}{2}\right)^2}{1.6 \times 10^{-8}} \)
\( = \frac{10 \times 3.14 \times 25}{4 \times 1.6} = 122.72 \text{ m} \)
\( \therefore \) length of the wire = 122.72m
If the diameter of the wire is doubled, new diameter=\( 2\times0.5 = 1\text{mm} = 0.001\text{m} \)
Let new resistance be R'
\( R' = \rho \frac{l}{A} \)
\( = \frac{1.6 \times 10^{-8} \times 122.72}{\pi (\frac{1}{2} \times 10^{-3})^2} \)
\( = \frac{1.6 \times 10^{-8} \times 122.72 \times 4}{3.14 \times 10^{-6}} \)
\( = 250.2 \times 10^{-2} = 2.5 \ \Omega \)
Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 \( \Omega \).

Question. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below −

Plot a graph between V and I and calculate the resistance of that resistor.
Answer: The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table.

The IV characteristic of the given resistor is plotted in the following figure.

The slope of the line gives the value of resistance (R) as,
Slope = 1/R = BC/AC = 2/6.8
R= 6.8/2 = 3.4 \( \Omega \)
Therefore, the resistance of the resistor is 3.4 \( \Omega \).

Question. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer: Resistance (R) of a resistor is given by Ohm's law as, V= IR
R= V/I
Where,
Potential difference, V= 12 V
Current in the circuit, I= 2.5 mA = 2.5 x 10-3 A
\[ R = \frac{12}{2.5 \times 10^{-3}} = 4.8 \times 10^3 \ \Omega = 4.8 \text{ k}\Omega \]
Therefore, the resistance of the resistor is 4.8 k\( \Omega \)

Question. A battery of 9 V is connected in series with resistors of 0.2 \( \Omega \), 0.3 \( \Omega \), 0.4 \( \Omega \), 0.5 \( \Omega \) and 12 \( \Omega \), respectively. How much current would flow through the 12 \( \Omega \) resistor?
Answer: There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as
V= IR
I= V/R
Where,
R is the equivalent resistance of resistances 0.2 \( \Omega \), 0.3 \( \Omega \), 0.4 \( \Omega \), 0.5 \( \Omega \) and 12 \( \Omega \). These are connected in series. Hence, the sum of the resistances will give the value of R.
R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \( \Omega \)
Potential difference, V= 9 V
I= 9/13.4 = 0.671 A
Therefore, the current that would flow through the 12 \( \Omega \) resistor is 0.671 A.

Question. How many 176 \( \Omega \) resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer: For x number of resistors of resistance 176 \( \Omega \), the equivalent resistance of the resistors connected in parallel is given by Ohm's law as V= IR
R= V/I
Where,
Supply voltage, V= 220 V
Current, I = 5 A
Equivalent resistance of the combination = R, given as
\[ \frac{1}{R} = x \times \left(\frac{1}{176}\right) \]
\[ R = \frac{176}{x} \]
From Ohm's law,
\[ \frac{V}{I} = \frac{176}{x} \]
\[ x = \frac{176 \times I}{V} = \frac{176 \times 5}{220} = 4 \]
Therefore, four resistors of 176 \( \Omega \) are required to draw the given amount of current.

Question. Show how you would connect three resistors, each of resistance 6 \( \Omega \), so that the combination has a resistance of (i) 9 \( \Omega \), (ii) 4 \( \Omega \).
Answer: If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 \( \Omega \) + 6 \( \Omega \) + 6 \( \Omega \) = 18 \( \Omega \), which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6/2 = 3 \( \Omega \) is also not desired. Hence, we should either connect the two resistors in series or parallel.
(a) Two resistors in parallel:
Two 6 \( \Omega \) resistors are connected in parallel. Their equivalent resistance will be
\[ \frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} \Rightarrow R_p = \frac{6 \times 6}{6 + 6} = 3 \ \Omega \]
The third 6 \( \Omega \) resistor is in series with 3 \( \Omega \). Hence, the equivalent resistance of the circuit is 6 \( \Omega \)+ 3 \( \Omega \) = 9 \( \Omega \).

(b) Two resistors in series:
Two 6 \( \Omega \) resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 \( \Omega \).
The third 6 \( \Omega \) resistor is in parallel with 12 \( \Omega \). Hence, equivalent resistance will be
\[ \frac{1}{R} = \frac{1}{12} + \frac{1}{6} \Rightarrow R = \frac{12 \times 6}{12 + 6} = 4 \ \Omega \]
Therefore, the total resistance is 4 \( \Omega \).

Question. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer: Resistance \( R_1 \) of the bulb is given by the expression,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P=10watts
Because \( R=V^2/P \)
\[ R_1 = \frac{(220)^2}{10} = 4840 \ \Omega \]
According to Ohm's law,
V= IR
Let R is the total resistance of the circuit for x number of electric bulbs
R=V/I
\[ R = \frac{220}{5} = 44 \ \Omega \]
Resistance of each electric bulb, \( R_1 = 4840 \ \Omega \)
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_1} + \dots \text{ upto } x \text{ times.} \]
\[ \frac{1}{R} = \frac{1}{R_1} \times x \]
\[ x = \frac{R_1}{R} = \frac{4840}{44} = 110 \]
\( \therefore \) Number of electric bulbs connected in parallel are 110.

Question. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 \( \Omega \) resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer: Supply voltage, V= 220 V
Resistance of one coil, R= 24 \( \Omega \)
(i) Coils are used separately
According to Ohm's law,
\( V= I_1R_1 \)
Where,
\( I_1 \) is the current flowing through the coil
\( I_1 = V/R_1 = 220/24 = 9.166 \text{ A} \)
Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series
Total resistance, \( R_2 = 24 \ \Omega + 24 \ \Omega = 48 \ \Omega \)
According to Ohm's law, V = \( I_2R_2 \)
Where,
\( I_2 \) is the current flowing through the series circuit
\( I_2 = V/R_2 = 220/48 = 4.58 \text{ A} \)
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.

(iii) Coils are connected in parallel
Total resistance, \( R_3 \) is given as =
\[ \frac{1}{R_3} = \frac{1}{24} + \frac{1}{24} \Rightarrow R_3 = 12 \ \Omega \]
According to Ohm's law,
V= \( I_3R_3 \)
Where,
\( I_3 \) is the current flowing through the circuit, \( I_3 = V/R_3 = 220/12 = 18.33 \text{ A} \)
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

Question. Compare the power used in the 2 \( \Omega \) resistor in each of the following circuits: (i) a 6 V battery in series with 1 \( \Omega \) and 2 \( \Omega \) resistors, and (ii) a 4 V battery in parallel with 12 \( \Omega \) and 2 \( \Omega \) resistors.
Answer: (i) Potential difference, V = 6 V
1 \( \Omega \) and 2 \( \Omega \) resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 \( \Omega \)
According to Ohm’s law,
V = IR
Where,
I is the current through the circuit
I= 6/3 = 2 A
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 \( \Omega \) resistor is 2 A.
Power is given by the expression,
\( P= (I)^2R = (2)^2 \times 2 = 8 \text{ W} \)

(ii) Potential difference, V = 4 V
12 \( \Omega \) and 2 \( \Omega \) resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 \( \Omega \) resistor will be 4 V.
Power consumed by 2 \( \Omega \) resistor is given by
\( P= V^2/R = 4^2/2 = 8 \text{ W} \)
Therefore, the power used by 2 \( \Omega \) resistor is 8 W.

Question. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer: Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by, Power = Voltage x Current
Current = Power/Voltage = 100/220 A
Current drawn by the bulb of rating 60 W is given by, Current = Power/Voltage = 60/220 A
Hence, current drawn from the line = 100/220 + 60/220 = 0.727 A

Question. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer: Energy consumed by an electrical appliance is given by the expression, H= Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = \( 250 \times 3600 = 9 \times 10^5 \text{ J} \)
Energy consumed by a toaster of power 1200 W in 10 minutes = \( 1200 \times 600 = 7.2 \times 10^5 \text{ J} \)
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

Question. An electric heater of resistance 8 \( \Omega \) draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer: Rate of heat produced by a device is given by the expression for power as, \( P= I^2R \)
Where,
Resistance of the electric heater, R= 8 \( \Omega \)
Current drawn, I = 15 A
P= \( (15)^2 \times 8 = 1800 \text{ J/s} \)
Therefore, heat is produced by the heater at the rate of 1800 J/s.

Question. Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer: (a) The melting point of Tungsten is very high and it has very high resistivity so it does not burn easily at a high temperature.
(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of metals which produces large amount of heat.
(c) In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e. when area of cross section increases the resistance decreases or vice versa.
(e) Copper and aluminium are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.