NCERT Solutions Class 10 Science Chapter 12 Electricity

NCERT Solutions for Class 10 Science Chapter 12 Electricity

NCERT Books Solutions Class 10 Science Chapter 12 Electricity

Class 10 is the first stepping stone for a student in the competitive world. With the introduction of the CBSE Board Exam for class 10 a few years back, this has become an important gateway for a student as based on the results of class 10th a student selects his future stream of Science, Commerce, or Arts.

NCERT Solutions Class 10 Science Chapter 12 ElectricityTakshila Learning provides you with detailed and well explained NCERT Solutions for Class 10 of each chapter of each subject for NCERT Class 10. These NCERT Solutions help you to easily understand every concept so that you can score high in your CBSE Class 10 Board Exams.

Below you can find the NCERT solution for Class 10th Science. You can get a Solution for the all-important question of “Ch 12 Science Class 10 Electricity”

Important formulae:

 

·       I(current) = Q(charge)/T(time)

·       V(potential difference) =W(work done)/Q(charge)

·       3-V(potential difference) =I(current)*R(resistance)

·       4-Resistivity=(R *A)/L, where R is resistance, A is area and L is length

·       5-H(heat) =I2Rt

·       6-P(power)=VI=I2r=V2/r

·       7 -F(force)=Q(charge)*E(electric field intensity)

·       8 -V=(K *Q)/R, where K is constant of proportionality of columb’s law, R is distance or radius.

·       9-E=(KQ)/R2

·       10-F=KQ1Q2/R2

 

{{Columb’s law}}

Note:- K=9*109 N/m

 

 

 

 

Page No. – 200

 

Q1. What does an electric circuit mean?

Ans: An electric circuit consists of electric devices, switching devices, source of electricity, etc. that are connected by conducting wires.

 

 

Q2. Define the unit of current.

Ans: 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.

 

 

Q3. Calculate the number of electrons constituting one coulomb of charge.

Ans: One electron possesses a charge of 1.6 × 10−19 C, i.e., 1.6 × 10−19 C of charge is contained in 1 electron.

 

∴ 1 C of charge is contained in

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.

 

Page No. – 202

 

Q1. Name a device that helps to maintain a potential difference across a conductor.

Ans: Any source of electricity like cell, battery, power supply, etc. can help to maintain a potential difference across a conductor.

 

 

Q2.What is meant by saying that the potential difference between two points is 1 V?

Ans: When 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.

 

 

Q3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Ans: The energy given to each coulomb of charge is equal to the amount of work required to move it. The amount of work is given by the expression,

 

Potential difference =

 

where, q = 1 C

Potential difference = 6 V

Work done

 

Therefore, 6 J of energy is given to each coulomb of charge passing through a battery of 6 V.

 

 

 

Page No. – 209

 

Q1. On what factors does the resistance of a conductor depend?

Ans: The resistance of a conductor depends upon the following factors:

  1. Length of the conductor
  2. Cross-sectional area of the conductor
  3. Material of the conductor
  4. Temperature of the conductor

 

 

Q2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Ans:

Resistance of a wire,

 

Where,

= Resistivity of the material of the wire

 

l = Length of the wire

 

A = Area of cross-section of the wire

Resistance is inversely proportional to the area of cross-section of the wire.

Hence, thicker the wire, lower is the resistance of the wire and vice-versa. So, current can flow more easily through a thick wire than a thin wire.

 

 

Q3. 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?

Ans: By Ohm’s law

V=IR

R remains constant so,

R=V/I

V gets halved

V→V/2

But R remains constant

so

R=(V/2)/x                    (x is the new current)

x=V/2R

I=V/R

substitute I in x equation

x=I/2

So when potential gets halved, current also gets halved.

 

 

Q4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Ans: The resistivity of an alloy is higher than the pure metal. Even at high temperatures, the alloys do not melt easily. So the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.

 

 

Q5. Use the data in Table 12.2 to answer the following −

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

Ans: a) Resistivity of iron =  10 X 10-8 Ω m

 

Resistivity of mercury = 94 X 10-8 Ω m

 

Resistivity of mercury is more than that of iron. So 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. So it is the best conductor.

 

Page No. – 213

 

Q1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Ans: Three cells of potential 2 V, each connected in series, is equivalent to a battery of potential 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors of resistances 5 Ω, 8 Ω and 12 Ω respectively connected in series and a battery of potential 6 V.

 

 

Q2. 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 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Ans: To measure the current flowing through the resistors, an ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the 12 Ω resistors, a voltmeter should be connected parallel to this resistor, 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 V

 

Current flowing through the circuit/resistors = I

 

Resistance of the circuit, R = 5 + 8 + 12 = 25 Ω

 

I =V/R=6/25 = 0.24 A

 

Potential difference across 12 Ω resistor = V1

 

Current flowing through the 12 Ω resistor, I = 0.24 A

 

By substituting Ohm’s law, we obtain

 

V1 = IR = 0.24 X 12 = 2.88 V

 

Therefore, the reading of the ammeter will be 0.24 A.

 

The reading of the voltmeter will be 2.88 V.

 

 

 

Page No. – 216

 

Q1. Judge the equivalent resistance when the following are connected in parallel –

(a) 1 Ω and 106Ω,

(b) 1 Ω and 103Ω and 106Ω.

Ans:

 

Q2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω 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?

Ans:Resistance of electric lamp (R1) = 100 Ω

Resistance of toaster (R2) = 5  0 Ω

Resistance of water filter (R3) = 500 Ω

Potential difference of the source, V = 220 V

As all the resistance are connected in parallel, the equivalent resistance R of the circuit will be

 

According to Ohm’s law

 

Where current flowing through the circuit = I

 

All the three given appliances are drawing 7.04 A of current.

Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A

Let   Rʹ be the resistance of the electric iron. According to Ohm’s law,

 

Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.

 

 

Q3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Ans: There is no division of voltage among the appliances when connected in parallel.In this case, 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.

 

Q4. How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Ans: There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.

The following circuit diagram shows the connection of the three resistors.

Here, 6 Ω and 3 Ω resistors are connected in parallel.

Here, 6 Ω and 3 Ω resistors are connected in parallel.

 

Therefore, their equivalent resistance will be given by

 

1/(1/6 + 1/3) = (6 X 3)/(6 + 3) = 2 Ω

 

 

This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.

 

Therefore, equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω.

 

Hence, the total resistance of the circuit is .

 

The following circuit diagram shows the connection of the three resistors.

 

All the resistors are connected in series. Therefore, their equivalent resistance will be given as

 

1/(1/2 + 1/3 + 1/6) =1/(3 + 2 + 1)/6 = 1 Ω

 

Therefore, the total resistance of the circuit is 1 Ω.

 

 

 

Q5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Ans: There are four coils of resistances, 4 Ω, 8 Ω, 12 Ω, and 24 Ω respectively.

(a) If these coils are connected in series, then the equivalent resistance will be the highest and will be given by the sum of resistances 4 + 8 + 12 + 24 = 48 Ω

 

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest and will be given by

 

1/(1/4 + 1/8 + 1/12 + 1/24) = 24/12 = 2 Ω

Therefore, 2 Ω is the lowest total resistance.

 

 

 

Page No. – 218

 

Q1. Why does the cord of an electric heater not glow while the heating element does?

Ans: The warming component of an electric warmer is a resistor. The measure of warmth created by it is relative to its opposition. The opposition of the component of an electric radiator is extremely high. As current courses through the warming component, it turns out to be excessively hot and sparkles red. Then again, the opposition of the rope is low. It doesn’t become red when current courses through it.

 

 

Q2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Ans: The amount of heat (H) produced is given by the Joule’s law of heating as

 

H = VIt

Where,

Voltage, V = 50 V

 

Time, t = 1 h = 1 × 60 × 60 s

 

Amount of current,

 

I = (Amount of charge)/(Time flow of charge)

 

= (96000)/(1 X 60 X 60) = 80/3 A

 

Therefore, the heat generated is .

 

H = 50 X (80/3) X 60 X 60 = 4.8 X 106 J.

 

 

 

Q3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Ans: 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 × Resistance = 5 × 20 = 100 V

 

H = 100 X 5 X 30 = 1.5 X 10 4 J

 

Therefore, the amount of heat developed in the electric iron is 1.5 X 10 4 J.

 

 

 

Page No. – 220

 

Q1. What determines the rate at which energy is delivered by a current?

Ans: The pace of utilization of electric vitality in an electric machine is called electric force. Henceforth, the rate at which vitality is conveyed by a current is the intensity of the machine.

 

 

Q2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Ans: Power (P) is given by the expression,

 

P = V I

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 × 60 × 60 = 7200 s

 

P = 1100 × 7200 = 7.92 × 106 J

 

Therefore, power of the motor = 1100 W

 

Energy consumed by the motor = 7.92 × 106 J

 

 

 

Page No. – 221

 

Q1. 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 −

  • 1/25
  • 1/5
  • 5
  • 25

 

Ans:  (d) Resistance of a piece of wire is proportional to its length. A piece of wire has a resistance R. The wire is cut into five equal parts.

 

Therefore, resistance of each part = R/5

 

All the five parts are connected in parallel. Hence, equivalent resistance (R’) is given as

 

1/R’ = 5/R + 5/R + 5/R + 5/R + 5/R =25/R

 

R’/R = 25

 

Therefore, the ratio is 25.

 

 

Q2. Which of the following terms does not represent electrical power in a circuit?

  • I2R
  • IR2
  • VI
  • V2/R

Ans: (b) Electrical power is given by the expression, P = VI … (i)

According to Ohm’s law, V = IR … (ii)

Where,

V = Potential difference

 

I = Current

 

R = Resistance

 

P = VI

 

From equation (i), it can be written

 

P = (IR) × I

 

P = I2R

 

From equation (ii), it can be written

 

I = V/R

 

P = V X V/R

 

P = V2/R

 

  • P = VI = I2/R = V2/R

 

Power P cannot be expressed as IR2.

 

Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be −

  • 100 W
  • 75 W
  • 50 W
  • 25 W

Ans: (d) Energy consumed by an appliance is given by the expression,

 

Where, P = VI = V2/R

 

R = V2/P

 

Power rating, P = 100 W

 

Voltage, V = 220 V

 

Resistance, R = (220)2/100 = 484 Ω

 

The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is calculated by the expression for power as

 

P’ = (V’)2  / R = (110)2/R = 25 W

 

Therefore, the power consumed will be 25 W.

 

 

Q4. 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 −

  • 1:2
  • 2:1
  • 1:4
  • 4:1

 

Ans: (c) The Joule heating is given by, H = i2Rt

 

Let, R be the resistance of the two wires.

 

The equivalent resistance of the series connection is RS = R + R = 2R

 

If V is the applied potential difference, then it is the voltage across the equivalent resistance.

 

V = iS X 2R

  • iS = V / 2R

 

The heat dissipated in time t is,

 

Hs = (iS)2 X 2R X t = (V/2R) 2 X 2R X t = V2t/2R

 

The equivalent resistance of the parallel connection is RP = 1/(1/R + 1/R) = R/2

 

V is the applied potential difference across this RP.

 

V = ip X (R/2)

  • ip = 2V/R

 

The heat dissipated in time t is,

 

Hp = (ip) 2 X R/2 X t = (2V/R)2 X R/2 X t

 

  • Hp = 2V2 t / R

 

So, the ratio of heat produced is,

 

Hs / Hp = (V2 t/2R)/(2V2t /R) = 1/4

 

 

Note:  H is proportional to  R

H is proportional to  i2

H is proportional to t.

In this problem, t is same in both the circuit.

But the current through the equivalent resistance of both the circuit is different.

As we know the voltage and resistance of the circuits, we have calculated i in terms of voltage and resistance and used in the equation H = i2Rt to find the ratio.

 

 

 

Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Ans: A voltmeter should be connected in parallel to the points, if we have to measure the potential difference between any two points.

 

 

 

Q6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Ans: Resistance (R) of a copper wire of length l and cross-section A is given by the expression,

R = ρ l/A

 

Where,

Resistivity of copper, ρ = 1.6 X 10-8 Ωm

Area of cross-section of the wire, A = π( Diameter /2 )2

Diameter= 0.5 mm = 0.0005 m

Resistance, R = 10 Ω

 

Hence, length of the wire,

l= RA/ρ = 122.72m

If the diameter of the wire is doubled, new diameter = 2X0.5 = 1 mm = 0.001 m

 

Therefore, resistance R’=

 

R’ = ρ l/A = 2.5 Ω

 

Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω.

 

 

 

Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

 

V (volts)

 

1.6

 

1.0

 

2.0

 

3.0

 

4.0

 

I (amperes )

 

0.5

 

3.4

 

6.7

 

10.2

 

13.2

 

Plot a graph between V and I and calculate the resistance of that resistor.

Ans: 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.

 

V (volts)

 

1.6

 

1.0

 

2.0

 

3.0

 

4.0

 

I (amperes )

 

0.5

 

3.4

 

6.7

 

10.2

 

13.2

 

 

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 = 3.4 Ω

 

Therefore, the resistance of the resistor is  3.4 Ω

 

 

 

Q8. 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.

Ans: 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 = 12 /(2.5 X 10  -3) = 4.8 X 103 Ω = 4.8 k Ω

 

Therefore, the resistance of the resistor is 4.8 k Ω.

 

 

 

Q9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Ans: No current division occurs in a series circuit.

Current flow through the component is given by Ohm’s law as

V = IR

R = V/I

 

Where,

R is the equivalent resistance of resistances. 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 Ω

 

Potential difference, V = 9 V

 

I = 9 /13.4 = 0.671 A

 

Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.

 

 

 

Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Ans: Let x be total number of resistors of resistance of 176 Ω each.

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

 

1/R = x X (1 /176)

 

R = 176 / x

From Ohm’s law,

 

V / I = 176/ x

 

x = 176 X I /V = 176 X 5/ 220 = 4

 

Therefore, four resistors of 176 Ω are required to draw the given amount of current.

 

 

 

Q11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Ans: If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be

6/2 = 3 Ω which is not desired.

 

Hence, we should either connect the two resistors in series or parallel.

 

(i) Two resistors in parallel

 

Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be

 

1 /(1/6 + 1/6) =  6 X 6 / 6 + 6 = 3 Ω

 

The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.

 

(ii) Two resistors in series

 

 

Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 Ω + 6 Ω = 12 Ω

 

The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be

 

1/(1/12 + 1/6) = 4 Ω

 

Therefore, the total resistance is  4 Ω.

 

 

 

Q12. 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?

Ans: Resistance R1 of the bulb is given by the expression,

Supply voltage, V = 220 V

Maximum current, I = 5 A

Rating of an electric bulb P = 10 watts

As R=V2/P

 

According to Ohm’s law,

V = I R

Let R is the total resistance of the circuit for x number of electric bulbs

R=V/I

=220/5=44 Ω

Resistance of each electric bulb, R1= 4840Ω

 

∴ Number of electric bulbs connected in parallel are 110.

 

 

 

Q13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Ans: Supply voltage, V = 220 V

 

Here Resistance of one coil, R =24 Ω

 

(i) Coils are used separately

 

According to Ohm’s law, V=I1R1

 

I1 is the current flowing through the coil

 

I1=V/R1=220/24=9.166A

 

Hence, current flow through the coil when used separately is 9.16 A.

 

(ii) Coils are connected in series

 

R2= 24 Ω + 24 Ω = 48 Ω

 

According to Ohm’s law

 

I2=V/R2=220/48=4.58A

 

So, the current flowing through the series circuit is 4.58A.

 

(iii) Coils are connected in parallel

 

According to Ohm’s law

 

I3=V/R3=220/12=18.33A

 

So, the current flowing through the parallel circuit is 18.33A.

 

 

 

Q14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Ans:

 

 

Q15. 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?

Ans: Both the bulbs are connected in parallel. So, the 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

 

Similarly, current drawn by the bulb of rating 60 W is given by,

 

Power = Voltage X Current

 

Current = Power / Voltage = 60/220 A

 

Hence, current drawn from line is

 

100/220 + 60/220 = 0.727 A

 

 

 

Q16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

 

Ans: Energy consumed by an electrical appliance is calculated 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 × 3600 = 9 × 105 J

 

Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 × 600

 

= 7.2× 105 J

 

Therefore, we can say that 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.

 

 

Q17. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Ans: Rate of heat produced by a device is calculated by the expression for power as

 

P = I2R

Where,

Resistance of the electric heater, R = 8 Ω

 

Current drawn, I = 15 A

 

P = (15) 2X 8 = 1800 J/s

 

Therefore, heat is produced by the heater at the rate of 1800 J/s.

 

 

 

Q18. Explain the following-

  • Why is the tungsten used almost exclusively for filament of electric lamps?
  • Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
  • Why is the series arrangement not used for domestic circuits?
  • How does the resistance of a wire vary with its area of cross-section?
  • Why copper and aluminum wires are usually employed for electricity transmission?

 

Ans: (a) The melting point and resistivity of tungsten are exceptionally high. It doesn’t burn promptly at a high temperature. The electric lights glow at extremely high temperatures. Thus, tungsten is basically utilized as heating component of electric bulbs.

 

(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy as resistivity of an alloy is more than that of metals.

 

(c) There is voltage division in series circuits. Each component of a series circuit receives a small voltage for a large supply voltage. As a result, the amount of current decreases and the device becomes hot. 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)

 

(e) Copper and aluminum wires have low resistivity. They are good conductors of electricity. Hence, they are usually employed for electricity transmission

 

 

 

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