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

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. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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.

According to Joule’s law of heating, the amount of heat produced is given by H = VIt Where, V = IR = 5A × 20 Ω = 100 V I = 5 A and t = 30 seconds So, 𝐻=100×5×30 𝐽 =15000 𝐽 =1.5×10⁴ 𝐽 For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

According to Joule’s law of heating, the amount of heat produced is given by
H = VIt
Where,
V = IR = 5A × 20 Ω = 100 V
I = 5 A
and t = 30 seconds
So, 𝐻=100×5×30 𝐽
=15000 𝐽
=1.5×10⁴ 𝐽

According to Joule’s law of heating, the amount of heat produced is given by H = VIt Where, V = 50 V 𝐼= 𝐶ℎ𝑎𝑟𝑔𝑒/𝑡𝑖𝑚𝑒= 9600 𝑐𝑜𝑢𝑙𝑜𝑚𝑏/1 ℎ𝑟 = 9600/60×60 = 803 𝐴 and t = 1 hour = 60 × 60 seconds So, 𝐻=50×80/3×60×60 =4800000 𝐽=4.8×10⁶ 𝐽 For more answers visit to website: https://www.tiwariacademy.com/ncertRead more

According to Joule’s law of heating, the amount of heat produced is given by
H = VIt
Where,
V = 50 V
𝐼= 𝐶ℎ𝑎𝑟𝑔𝑒/𝑡𝑖𝑚𝑒= 9600 𝑐𝑜𝑢𝑙𝑜𝑚𝑏/1 ℎ𝑟 = 9600/60×60 = 803 𝐴
and t = 1 hour = 60 × 60 seconds
So,
𝐻=50×80/3×60×60
=4800000 𝐽=4.8×10⁶ 𝐽

The heating element of an electric heater is a resistor. According to Joule’s law of heating, the amount of heat produced by it is proportional to its resistance. H = I²Rt The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hRead more

The heating element of an electric heater is a resistor. According to Joule’s law of heating, the amount of heat produced by it is proportional to its resistance.
H = I²Rt
The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hot and glows red. On the other hand, the resistance of the cord is low. It does not become red when current flows through it.

Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance. (a) The highest total resistance is given by 𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄ =4 Ω+ 8 Ω+ 12 Ω+ 24 Ω =48 Ω (b) The lowest total resistance is given by 1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄ 1𝑅=1/4+1/8+1/12+1/24 = 6+3+2+1/24 = 12/2Read more

Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance.
(a) The highest total resistance is given by
𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄
=4 Ω+ 8 Ω+ 12 Ω+ 24 Ω
=48 Ω
(b) The lowest total resistance is given by
1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄
1𝑅=1/4+1/8+1/12+1/24
= 6+3+2+1/24 = 12/24
⟹𝑅=24/12=2 Ω

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.

Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel. The net resistance in parallel is given by 1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃ Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So, 1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.2Read more

Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel.
The net resistance in parallel is given by
1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃
Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So,
1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.25 Ω
Now, using Ohm’s law V = IR, we have
𝐼=𝑉/𝑅 = 220 𝑉/31.25 Ω = 7.04 𝐴
Hence, the resistance of electric iron is 31.25 Ω and current through it is 7.04 A.

(a) The net resistance in parallel is given by 1/𝑅=1/𝑅₁ +1/𝑅₂ Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So, 1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω (b) The net resistance in parallel is given by 1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃ Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So, 1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1Read more

(a) The net resistance in parallel is given by
1/𝑅=1/𝑅₁ +1/𝑅₂
Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So,
1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω
(b) The net resistance in parallel is given by
1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃
Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So,
1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1000000 ⟹𝑅=1000000/1001001= 0.999Ω ≈ 1Ω

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.

## 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 × 5 = 1100 W Now, the energy consumed = Power × time Where, P = 1100 W t = 2 hours = 2 × 60 × 60 seconds = 7200 seconds So, the energy consumed E = 1100 × 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 × 5 = 1100 W

Now, the energy consumed = Power × time

Where, P = 1100 W

t = 2 hours = 2 × 60 × 60 seconds = 7200 seconds

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

For more answers visit to website:

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

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

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. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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.

For more answers visit to website:

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

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

According to Joule’s law of heating, the amount of heat produced is given by H = VIt Where, V = IR = 5A × 20 Ω = 100 V I = 5 A and t = 30 seconds So, 𝐻=100×5×30 𝐽 =15000 𝐽 =1.5×10⁴ 𝐽 For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

According to Joule’s law of heating, the amount of heat produced is given by

H = VIt

Where,

V = IR = 5A × 20 Ω = 100 V

I = 5 A

and t = 30 seconds

So, 𝐻=100×5×30 𝐽

=15000 𝐽

=1.5×10⁴ 𝐽

For more answers visit to website:

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

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

According to Joule’s law of heating, the amount of heat produced is given by H = VIt Where, V = 50 V 𝐼= 𝐶ℎ𝑎𝑟𝑔𝑒/𝑡𝑖𝑚𝑒= 9600 𝑐𝑜𝑢𝑙𝑜𝑚𝑏/1 ℎ𝑟 = 9600/60×60 = 803 𝐴 and t = 1 hour = 60 × 60 seconds So, 𝐻=50×80/3×60×60 =4800000 𝐽=4.8×10⁶ 𝐽 For more answers visit to website: https://www.tiwariacademy.com/ncertRead more

According to Joule’s law of heating, the amount of heat produced is given by

H = VIt

Where,

V = 50 V

𝐼= 𝐶ℎ𝑎𝑟𝑔𝑒/𝑡𝑖𝑚𝑒= 9600 𝑐𝑜𝑢𝑙𝑜𝑚𝑏/1 ℎ𝑟 = 9600/60×60 = 803 𝐴

and t = 1 hour = 60 × 60 seconds

So,

𝐻=50×80/3×60×60

=4800000 𝐽=4.8×10⁶ 𝐽

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

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

The heating element of an electric heater is a resistor. According to Joule’s law of heating, the amount of heat produced by it is proportional to its resistance. H = I²Rt The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hRead more

The heating element of an electric heater is a resistor. According to Joule’s law of heating, the amount of heat produced by it is proportional to its resistance.

H = I²Rt

The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hot and glows red. On the other hand, the resistance of the cord is low. It does not become red when current flows through it.

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

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

Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance. (a) The highest total resistance is given by 𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄ =4 Ω+ 8 Ω+ 12 Ω+ 24 Ω =48 Ω (b) The lowest total resistance is given by 1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄ 1𝑅=1/4+1/8+1/12+1/24 = 6+3+2+1/24 = 12/2Read more

Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance.

See less(a) The highest total resistance is given by

𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄

=4 Ω+ 8 Ω+ 12 Ω+ 24 Ω

=48 Ω

(b) The lowest total resistance is given by

1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄

1𝑅=1/4+1/8+1/12+1/24

= 6+3+2+1/24 = 12/24

⟹𝑅=24/12=2 Ω

## 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.

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

## An electric lamp of 100 ohm, a toaster of resistance 50 ohm, and a water filter of resistance 500 ohm 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?

Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel. The net resistance in parallel is given by 1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃ Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So, 1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.2Read more

Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel.

The net resistance in parallel is given by

1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃

Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So,

1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.25 Ω

Now, using Ohm’s law V = IR, we have

𝐼=𝑉/𝑅 = 220 𝑉/31.25 Ω = 7.04 𝐴

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

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

## Judge the equivalent resistance when the following are connected in parallel

(a) The net resistance in parallel is given by 1/𝑅=1/𝑅₁ +1/𝑅₂ Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So, 1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω (b) The net resistance in parallel is given by 1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃ Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So, 1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1Read more

(a) The net resistance in parallel is given by

1/𝑅=1/𝑅₁ +1/𝑅₂

Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So,

1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω

(b) The net resistance in parallel is given by

1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃

Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So,

1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1000000 ⟹𝑅=1000000/1001001= 0.999Ω ≈ 1Ω

https://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.

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