Ohm's Law

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August 2nd 2007, 07:39 AM
Jonboy

Ohm's Law

Hi everyone again. These problems are tough... I need help... ¡Necesito 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.
August 2nd 2007, 07:41 AM
topsquark

Quote:

Originally Posted by Jonboy

Hi everyone again. These problems are tough... I need help... ¡Necesito 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
August 2nd 2007, 07:53 AM
Jonboy

Thanks, shouldn't I plug V = 110 and R = 50 in that final equation you got?
August 2nd 2007, 08:02 AM
topsquark

Quote:

Originally Posted by Jonboy

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