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In an electric circuit, two resistors with resistances x and

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In an electric circuit, two resistors with resistances x and [#permalink] New post 20 Aug 2008, 21:33
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In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the sum of the reciprocals of x and y. What is r in terms of x and y?

(a) xy

(b) x+y

(c) 1/ x+y

(d) xy/ x+y

(e) x+y / xy

How would you solve by plugging in? and algebra?
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Re: PS: reciprocal (OG #230) [#permalink] New post 20 Aug 2008, 22:05
its simple ..

1/r=1/x+1/y
1/r=x+y/xy
r=xy/x+y
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Re: PS: reciprocal (OG #230) [#permalink] New post 21 Aug 2008, 10:51
The beginning of question is confusing...

two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors


Does this have anything to do w/ the questions b/c to me, it seems the latter is purpose of question..
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Re: PS: reciprocal (OG #230) [#permalink] New post 21 Aug 2008, 11:01
droopy57 wrote:
The beginning of question is confusing...

two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors


Does this have anything to do w/ the questions b/c to me, it seems the latter is purpose of question..


It just means r = value of (x//y)
Basically... this is EE101 of finding the equivalent resistor value. :-D
1/r = 1/x + 1/y is the starting equation
1/r = (x+y)/xy
and r = xy/(x+y)
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Re: PS: reciprocal (OG #230) [#permalink] New post 21 Aug 2008, 11:08
That's great for all of those EE's out there...oh wait, they're actually a small % of MBA seekers.

For the rest of us

reciprocal of x = 1/x
reciprocal of r = 1/r
reciprocal of y = 1/y

1/r = 1/x + 1/y

To get the same denominator
\frac{1}{r} = \frac{y}{y}*\frac{1}{x} + \frac{x}{x}*\frac{1}{y}

becomes

\frac{1}{r} = \frac{y}{xy} + \frac{x}{xy}

\frac{1}{r} = \frac{X + y}{xy}

Now you can get rid of the fraction with r by cross multiplying
r * (x+y) = 1*xy
and then dividing to get r alone
r = xy/x+y
fatb wrote:
droopy57 wrote:
The beginning of question is confusing...

two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors


Does this have anything to do w/ the questions b/c to me, it seems the latter is purpose of question..


It just means r = value of (x//y)
Basically... this is EE101 of finding the equivalent resistor value. :-D
1/r = 1/x + 1/y is the starting equation
1/r = (x+y)/xy
and r = xy/(x+y)

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Manager
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Joined: 27 Mar 2008
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Re: PS: reciprocal (OG #230) [#permalink] New post 21 Aug 2008, 11:51
ok. i see where i went wrong. i did this:

1/r = 2/ x+y

wasn't thinking that i couldnt combine unless it was like terms...
Re: PS: reciprocal (OG #230)   [#permalink] 21 Aug 2008, 11:51
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