# Class 10 Science Chapter 12 Electricity

Class 10 Science Chapter 12 Electricity The answer to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 10 Science Chapter 12 Electricity and select need one.

## Class 10 Science Chapter 12 Electricity

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. These solutions are part of SCERT All Subject Solutions. Here we have given Assam Board Class 10 Science Chapter 12 Electricity Solutions for All Subjects, You can practice these here…

### Electricity

Chapter – 12

GENERAL SCIENCE

Page – 200

Q.1. What does an electric circuit mean?

Ans :- A continuous and closed path of an electric current is called an electric circuit.

Q.2. Define the unit of current?

Ans :- The unit of current is ampere. Flow of one coulomb of charge in one second is called one ampere.

i.e. 1 amp = 1 coulomb/1 second

Q.3. Calculate the number of electrons constituting one coulomb of charge.

Ans :- Charge on one electron (e) = 1.6×10⁻¹⁹C

Total charge Q  = 1C

∴     Number of electrons,  n  = Q/C

= 1C/1.6×10⁻¹⁹C

= 6.25×10¹⁸

Page – 202

Q.1. Name a device that helps to maintain a potential duo difference across a conductor.

Ans :- Battery.

Q.2. What is meant by saying that the potential difference between two points is 1V?

Ans :- One volt is the potential difference between two points in a current carrying conductor when 1 Joule of work is done to move a change of 1 coulomb from one point to the other.

Therefore 1 volt = 1 Joule/1 coulomb

Q.3. How much energy is given to each coulomb of charge passing through a 6v battery?

Ans :- Required energy = Charge x potential difference

= 1c x 6v

= 6 Joule

Page -209

Q.1. On what factors does the resistance of a conductor depend?

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

( i ) Length of the conductor.

( ii ) Area of cross section of the conductor.

( iii ) Nature of its material.

Q.2. Will current flow more easily through a thick wire or a thin wire the same material. When connected to the same source? Why?

Ans :- The current will flow more easily through a thick wire than a thin wire of the same material. Because resistance of the conductor is inversely proportional to the area of cross-section.

Q.3. Let the resistance of an electrical component remains constant while the potential difference across the two Isikn ends of the component decreases to half of its for men value. What change will occur in the current through it?

Ans :- When the potential difference across the two ends of the component decreases to half then the current through it also decreases to half of its initial value. Because V ∞ I.

Q.4. 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 generally higher than that of its constituent metals. Alloys do not oxidise readily at high temperatures. For this reason they are commonly used in electrical toasters and electric iron.

Q.5. Use the date 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 = 10.0 x 10⁻⁸Ωm

Thus iron is a better conductor because it has lower resistivity than mercury.

( b ) As silver has the lowest resistivity, so silver is the best conductor.

Page – 213

Q.1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2v each, a 5Ω resistor, and 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series.

Ans :- The circuit diagram is given below :-

Q.2. Redraw the circuit of questions 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and voltmeter? Ans. The circuit diagram is given below :

Ans :- The circuit diagram is given below :-

Total voltage,     V = 3×2 = 6v

Total resistance, R = 5+8+12 = 25 Ω

Reading of ammeter, I = V/R

= 6/25 = 0.24A

Reading of voltmeter, V =  I.R

=  0.24×12

=  2.88V

∴           =  2 Ω

∴       Total resistance = 2

Page – 218

Q.1. Why does the cord of an electric heater not glow while the heating element does?

Ans :- Both the cord and the heating element of an electric heater carry the same current. But the heating element becomes not due to its high resistance and begins to glow. The cord remains cold due to its low resistance and does not glow.

Q.2. Compute the heat generated while transferring 96,000 coulomb of charge in one hour through a potential difference of 50v.

Ans :- Here     Q   =   96,000c

t    =   1 hour

=    36 sec

V   =    50 V

Heat generated is     H  =  VQ

=  50×96,000J

=  48,00000J

Q.3. An electric iron of resistance 20 Ω tkes a current of 5A. Calculate the heat developed in 30 Sec.

Ans :- Here,    R   =  20 Ω

I    =  5A

t    =  30 sec

∴  Heat developed  =  I²Rt

= 5²x20x30 J

= 25x20x30 J

= 15000 J

= 15 KJ

Page – 220

Q.1. What determines the rate at which the energy is delivered by a current?

Ans :- The rate at which the energy is delivered by a current is called power.

The power P is given by   P   =   VI

or  P   =   I²R

= v²/R

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

Ans :- Here        I    =    5A

V   =    220v

P   =    ?

We have    P   =   VI

=   220 x 5

= 1100 w

Energy consumed  =  P x t

=  1100w x 2h

=  1100w x 2 x 60 x 60 sec

=  1100w x 7200 sec

=  7920000 J

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