Power of the electric motor is given by P = VI Where, V = 220 V and I = 5 A So, Power P = 220 Γ 5 = 1100 W Now, the energy consumed = Power Γ time Where, P = 1100 W t = 2 hours = 2 Γ 60 Γ 60 seconds = 7200 seconds So, the energy consumed E = 1100 Γ 7200 J = 7920000 J For more answers visit to websitRead more

Power of the electric motor is given by
P = VI
Where, V = 220 V and I = 5 A
So, Power P = 220 Γ 5 = 1100 W
Now, the energy consumed = Power Γ time
Where, P = 1100 W
t = 2 hours = 2 Γ 60 Γ 60 seconds = 7200 seconds
So, the energy consumed E = 1100 Γ 7200 J = 7920000 J

In parallel there is no division of voltage among the appliances. The potential difference across each appliance is equal to the supplied voltage and the total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel. For more answers visit to website: httpsRead more

In parallel there is no division of voltage among the appliances. The potential difference across each appliance is equal to the supplied voltage and the total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even 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. For more aRead more

The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even 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.

According to the Ohmβs law V = IR If the resistance remains constant, V is directly proportional to I. Therefore, πβπΌ Now, if potential difference is reduced to half of its value, the current also become half of its original value. For more answers visit to website: https://www.tiwariacademy.com/nceRead more

According to the Ohmβs law V = IR
If the resistance remains constant, V is directly proportional to I. Therefore, πβπΌ
Now, if potential difference is reduced to half of its value, the current also become half of its original value.

The resistance of a conductor depends upon the following factors: β’ Length of the conductor β’ Cross-sectional area of the conductor β’ Material of the conductor β’ Temperature of the conductor. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12Read more

The resistance of a conductor depends upon the following factors:
β’ Length of the conductor
β’ Cross-sectional area of the conductor
β’ Material of the conductor
β’ Temperature of the conductor.

πππ‘πππ‘πππ π·πππππππππ=ππππ ππππ/πΆβππππ or Work done (or Energy) = Potential Difference Γ Charge So, Work done = 6 Volt Γ 1 Coulomb = 6 Joules For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

πππ‘πππ‘πππ π·πππππππππ=ππππ ππππ/πΆβππππ
or Work done (or Energy) = Potential Difference Γ Charge
So, Work done = 6 Volt Γ 1 Coulomb = 6 Joules

When 1 J of work is required to move a charge of 1 C from one point to another, then it is said that the potential difference between the two points is 1 V. V=W/Q 1 V = 1 J/1 C For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

When 1 J of work is required to move a charge of 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.
V=W/Q
1 V = 1 J/1 C

A cell, battery, power supply, etc. helps to maintain a potential difference across a conductor. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

A cell, battery, power supply, etc. helps to maintain a potential difference across a conductor.

We know that one electron possesses a charge of 1.6 Γ 10β»ΒΉβΉ C. ππ’ππππ ππ πππππ‘πππ=πππ‘ππ πβππππ/πΆβππππ ππ 1 πππππ‘πππ =1/1.6Γ10β»ΒΉβΉ=6.25Γ10ΒΉβΈ So, the number of electrons constituting one coulomb of charge is 6Γ10ΒΉβΈ. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-Read more

We know that one electron possesses a charge of 1.6 Γ 10β»ΒΉβΉ C.
ππ’ππππ ππ πππππ‘πππ=πππ‘ππ πβππππ/πΆβππππ ππ 1 πππππ‘πππ
=1/1.6Γ10β»ΒΉβΉ=6.25Γ10ΒΉβΈ
So, the number of electrons constituting one coulomb of charge is 6Γ10ΒΉβΈ.

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

Power of the electric motor is given by P = VI Where, V = 220 V and I = 5 A So, Power P = 220 Γ 5 = 1100 W Now, the energy consumed = Power Γ time Where, P = 1100 W t = 2 hours = 2 Γ 60 Γ 60 seconds = 7200 seconds So, the energy consumed E = 1100 Γ 7200 J = 7920000 J For more answers visit to websitRead more

Power of the electric motor is given by

P = VI

Where, V = 220 V and I = 5 A

So, Power P = 220 Γ 5 = 1100 W

Now, the energy consumed = Power Γ time

Where, P = 1100 W

t = 2 hours = 2 Γ 60 Γ 60 seconds = 7200 seconds

So, the energy consumed E = 1100 Γ 7200 J = 7920000 J

For more answers visit to website:

See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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

In parallel there is no division of voltage among the appliances. The potential difference across each appliance is equal to the supplied voltage and the total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel. For more answers visit to website: httpsRead more

In parallel there is no division of voltage among the appliances. The potential difference across each appliance is equal to the supplied voltage and the total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

For more answers visit to website:

See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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

The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even 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. For more aRead more

The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even 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.

For more answers visit to website:

See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

## 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?

According to the Ohmβs law V = IR If the resistance remains constant, V is directly proportional to I. Therefore, πβπΌ Now, if potential difference is reduced to half of its value, the current also become half of its original value. For more answers visit to website: https://www.tiwariacademy.com/nceRead more

According to the Ohmβs law V = IR

If the resistance remains constant, V is directly proportional to I. Therefore, πβπΌ

Now, if potential difference is reduced to half of its value, the current also become half of its original value.

https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

## On what factors does the resistance of a conductor depend?

The resistance of a conductor depends upon the following factors: β’ Length of the conductor β’ Cross-sectional area of the conductor β’ Material of the conductor β’ Temperature of the conductor. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12Read more

The resistance of a conductor depends upon the following factors:

β’ Length of the conductor

β’ Cross-sectional area of the conductor

β’ Material of the conductor

β’ Temperature of the conductor.

https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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

πππ‘πππ‘πππ π·πππππππππ=ππππ ππππ/πΆβππππ or Work done (or Energy) = Potential Difference Γ Charge So, Work done = 6 Volt Γ 1 Coulomb = 6 Joules For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

πππ‘πππ‘πππ π·πππππππππ=ππππ ππππ/πΆβππππ

or Work done (or Energy) = Potential Difference Γ Charge

So, Work done = 6 Volt Γ 1 Coulomb = 6 Joules

https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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

When 1 J of work is required to move a charge of 1 C from one point to another, then it is said that the potential difference between the two points is 1 V. V=W/Q 1 V = 1 J/1 C For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

When 1 J of work is required to move a charge of 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.

V=W/Q

1 V = 1 J/1 C

https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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

A cell, battery, power supply, etc. helps to maintain a potential difference across a conductor. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

A cell, battery, power supply, etc. helps to maintain a potential difference across a conductor.

https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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

We know that one electron possesses a charge of 1.6 Γ 10β»ΒΉβΉ C. ππ’ππππ ππ πππππ‘πππ=πππ‘ππ πβππππ/πΆβππππ ππ 1 πππππ‘πππ =1/1.6Γ10β»ΒΉβΉ=6.25Γ10ΒΉβΈ So, the number of electrons constituting one coulomb of charge is 6Γ10ΒΉβΈ. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-Read more

We know that one electron possesses a charge of 1.6 Γ 10β»ΒΉβΉ C.

ππ’ππππ ππ πππππ‘πππ=πππ‘ππ πβππππ/πΆβππππ ππ 1 πππππ‘πππ

=1/1.6Γ10β»ΒΉβΉ=6.25Γ10ΒΉβΈ

So, the number of electrons constituting one coulomb of charge is 6Γ10ΒΉβΈ.

https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

## Compare the power used in the 2 ohm resistor in each of the following circuits:

Given that: Potential difference, V = 6 V (i) 1 Ξ© and 2 Ξ© resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ξ© According to Ohmβs law, V = IR βΉπΌ=π/π =6/3=2 π΄ In series combination, the current in the circuit remains constant. Therefore power is given by π=πΌΒ²π =(Read more

Given that: Potential difference, V = 6 V

(i) 1 Ξ© and 2 Ξ© resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ξ©

According to Ohmβs law, V = IR

βΉπΌ=π/π =6/3=2 π΄

In series combination, the current in the circuit remains constant. Therefore power is given by

π=πΌΒ²π =(2)Β²Γ2=8 π

(ii) 1 Ξ© and 2 Ξ© resistors are connected in parallel. βΉπΌ=π/π =6/3=2 π΄

In parallel combination, the voltage in the circuit remains constant. Therefore, power is given by π=πΒ²/π =4Β²/2=8 π

Hence, in both the cases power remains same as 8W.

https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/