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Thread: 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 $\displaystyle I\,=\,\frac{V}{R}$, where $\displaystyle I$ is the current in amperes, $\displaystyle V$ is the electromotive force in volts, and $\displaystyle R$ 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 $\displaystyle I$ to double?

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

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

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

    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 $\displaystyle I\,=\,\frac{V}{R}$, where $\displaystyle I$ is the current in amperes, $\displaystyle V$ is the electromotive force in volts, and $\displaystyle R$ 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 $\displaystyle I$ to double?

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

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

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

    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
    $\displaystyle I = \frac{V}{R}$

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

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

    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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    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
    $\displaystyle x = \frac{RV}{R - 2V}$

    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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