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Consult From physics

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  1. Asked: October 20, 2020In: Class 10

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

    Consult From physics
    Added an answer on December 6, 2020 at 7:38 am

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

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  2. Asked: October 20, 2020In: Class 10

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

    Consult From physics
    Added an answer on December 6, 2020 at 7:33 am

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

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  3. Asked: October 20, 2020In: Class 10

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

    Consult From physics
    Added an answer on December 6, 2020 at 7:27 am

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

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  4. Asked: October 20, 2020In: Class 10

    Calculate the number of electrons constituting one coulomb of charge.

    Consult From physics
    Added an answer on December 6, 2020 at 7:24 am

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

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  5. Asked: October 20, 2020In: Class 10

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

    Consult From physics
    Added an answer on December 6, 2020 at 6:53 am

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

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