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.
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/