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Math Help - Ohm's Law

  1. #1
    Member Jonboy's Avatar
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    Ohm's Law

    Hi everyone again. These problems are tough... I need help... íNecesito Ayuda!

    In electrical theory, Ohm's law states that I\,=\,\frac{V}{R}, where I is the current in amperes, V is the electromotive force in volts, and R is the resistance in Ohms.
    In a certain circuit V = 110 and R = 50. If V and R are to be changed by the same numerical amount, what change in them will cause I to double?

    First I thought it would be imperative to find the new doubled amount.

    So currently we have: I\,=\,\frac{110}{50}\,=\,\frac{11}{5}

    Double that would be \frac{22}{10}

    I don't see how to make an equation to increase both the V and R by a constant number. Please help and thanks everyone.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jonboy View Post
    Hi everyone again. These problems are tough... I need help... íNecesito Ayuda!

    In electrical theory, Ohm's law states that I\,=\,\frac{V}{R}, where I is the current in amperes, V is the electromotive force in volts, and R is the resistance in Ohms.
    In a certain circuit V = 110 and R = 50. If V and R are to be changed by the same numerical amount, what change in them will cause I to double?

    First I thought it would be imperative to find the new doubled amount.

    So currently we have: I\,=\,\frac{110}{50}\,=\,\frac{11}{5}

    Double that would be \frac{22}{10}

    I don't see how to make an equation to increase both the V and R by a constant number. Please help and thanks everyone.
    We know that
    I = \frac{V}{R}

    So let
    2I = \frac{V + x}{R + x}

    Putting the first equation into the second:
    2 \frac{V}{R} = \frac{V + x}{R + x}

    Solve this for x.

    -Dan
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  3. #3
    Member Jonboy's Avatar
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    Thanks, shouldn't I plug V = 110 and R = 50 in that final equation you got?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jonboy View Post
    Thanks, shouldn't I plug V = 110 and R = 50 in that final equation you got?
    It's better practice to solve for x in terms of V and R, but if that's the way you want to do it, go for it!

    For reference I get
    x = \frac{RV}{R - 2V}

    Edit: Oh dear. That makes x negative. Either my setup is wrong or there is no solution.

    Edit Mark II: No, a negative x is okay because the questions says that we change V and R by the same amount, not add.

    -Dan
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