1. Given, Charge Q = 1C, Potential difference, V = 6V Therefore, Energy i.e. Work done, W =? We know that, V = W/Q Therefore, 6V = W/1C ⇒ W = 6V × 1C = 6J Thus, required energy = 6J

    Given, Charge Q = 1C, Potential difference, V = 6V

    Therefore, Energy i.e. Work done, W =?
    We know that, V = W/Q
    Therefore, 6V = W/1C
     W = 6V × 1C = 6J
    Thus, required energy = 6J

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  2. This means 1 joule of work is done to move a charge of 1 coulomb between two points.

    This means 1 joule of work is done to move a charge of 1 coulomb between two points.

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  3. We know that charge over 1 electron = 1.6 × 10–19 coulomb Thus, 1.6 × 10–19 C of charge = 1 electron Therefore, 1 C of charge = 1/ (1.6 × 10–19) Electrons = 1019/1.6 electrons = 10×1018/1.6 electrons = 6.25 × 1018 electrons

    We know that charge over 1 electron = 1.6 × 10–19 coulomb

    Thus, 1.6 × 10–19 C of charge = 1 electron
    Therefore, 1 C of charge = 1/ (1.6 × 10–19) Electrons
    = 1019/1.6 electrons = 10×1018/1.6 electrons = 6.25 × 1018 electrons

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  4. SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere. Therefore, 1 ampere = 1C/1s

    SI unit of electric current is ampere (A).

    Ampere is the flow of electric charges through a surface at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.
    Therefore, 1 ampere = 1C/1s

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