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
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 Ohm’s law V = IR If the resistance remains constant, V is directly proportional to I. Therefore, 𝑉∝𝐼 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 Ohm’s law V = IR
If the resistance remains constant, V is directly proportional to I. Therefore, 𝑉∝𝐼
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: • Length of the conductor • Cross-sectional area of the conductor • Material of the conductor • 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:
• Length of the conductor
• Cross-sectional area of the conductor
• Material of the conductor
• Temperature of the conductor.
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒=𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒/𝐶ℎ𝑎𝑟𝑔𝑒 or Work done (or Energy) = Potential Difference × Charge So, Work done = 6 Volt × 1 Coulomb = 6 Joules For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒=𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒/𝐶ℎ𝑎𝑟𝑔𝑒
or Work done (or Energy) = Potential Difference × Charge
So, Work done = 6 Volt × 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 × 10⁻¹⁹ C. 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛=𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒/𝐶ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 =1/1.6×10⁻¹⁹=6.25×10¹⁸ So, the number of electrons constituting one coulomb of charge is 6×10¹⁸. 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 × 10⁻¹⁹ C.
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛=𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒/𝐶ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
=1/1.6×10⁻¹⁹=6.25×10¹⁸
So, the number of electrons constituting one coulomb of charge is 6×10¹⁸.
Given that: Potential difference, V = 6 V (i) 1 Ω and 2 Ω resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω According to Ohm’s law, V = IR ⟹𝐼=𝑉/𝑅=6/3=2 𝐴 In series combination, the current in the circuit remains constant. Therefore power is given by 𝑃=𝐼²𝑅=(Read more
Given that: Potential difference, V = 6 V
(i) 1 Ω and 2 Ω resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law, V = IR
⟹𝐼=𝑉/𝑅=6/3=2 𝐴
In series combination, the current in the circuit remains constant. Therefore power is given by
𝑃=𝐼²𝑅=(2)²×2=8 𝑊
(ii) 1 Ω and 2 Ω resistors are connected in parallel. ⟹𝐼=𝑉/𝑅=6/3=2 𝐴
In parallel combination, the voltage in the circuit remains constant. Therefore, power is given by 𝑃=𝑉²/𝑅 =4²/2=8 𝑊
Hence, in both the cases power remains same as 8W.
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 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 Ohm’s law V = IR If the resistance remains constant, V is directly proportional to I. Therefore, 𝑉∝𝐼 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 Ohm’s law V = IR
If the resistance remains constant, V is directly proportional to I. Therefore, 𝑉∝𝐼
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:
See lesshttps://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: • Length of the conductor • Cross-sectional area of the conductor • Material of the conductor • 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:
• Length of the conductor
• Cross-sectional area of the conductor
• Material of the conductor
• Temperature of the conductor.
For more answers visit to website:
See lesshttps://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?
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒=𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒/𝐶ℎ𝑎𝑟𝑔𝑒 or Work done (or Energy) = Potential Difference × Charge So, Work done = 6 Volt × 1 Coulomb = 6 Joules For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒=𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒/𝐶ℎ𝑎𝑟𝑔𝑒
or Work done (or Energy) = Potential Difference × Charge
So, Work done = 6 Volt × 1 Coulomb = 6 Joules
For more answers visit to website:
See lesshttps://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
For more answers visit to website:
See lesshttps://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.
For more answers visit to website:
See lesshttps://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 × 10⁻¹⁹ C. 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛=𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒/𝐶ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 =1/1.6×10⁻¹⁹=6.25×10¹⁸ So, the number of electrons constituting one coulomb of charge is 6×10¹⁸. 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 × 10⁻¹⁹ C.
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛=𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒/𝐶ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
=1/1.6×10⁻¹⁹=6.25×10¹⁸
So, the number of electrons constituting one coulomb of charge is 6×10¹⁸.
For more answers visit to website:
See lesshttps://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 Ω and 2 Ω resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω According to Ohm’s law, V = IR ⟹𝐼=𝑉/𝑅=6/3=2 𝐴 In series combination, the current in the circuit remains constant. Therefore power is given by 𝑃=𝐼²𝑅=(Read more
Given that: Potential difference, V = 6 V
(i) 1 Ω and 2 Ω resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law, V = IR
⟹𝐼=𝑉/𝑅=6/3=2 𝐴
In series combination, the current in the circuit remains constant. Therefore power is given by
𝑃=𝐼²𝑅=(2)²×2=8 𝑊
(ii) 1 Ω and 2 Ω resistors are connected in parallel. ⟹𝐼=𝑉/𝑅=6/3=2 𝐴
In parallel combination, the voltage in the circuit remains constant. Therefore, power is given by 𝑃=𝑉²/𝑅 =4²/2=8 𝑊
Hence, in both the cases power remains same as 8W.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/