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# In an electric circuit, two resistors with resistances x and

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Manager
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In an electric circuit, two resistors with resistances x and [#permalink]  29 Dec 2012, 04:39
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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
[Reveal] Spoiler: OA
Math Expert
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Re: In an electric circuit, two resistors with resistances x and [#permalink]  28 Dec 2013, 02:42
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aeglorre wrote:
Bunuel wrote:
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

The wording is a bit confusing, though basically we are told that 1/r = 1/x + 1/y, from which it follows that r=xy/(x + y).

Could you explain this more in depth?

I interpret it as:

" r is the combined of x and y" --> r = x + y

"the reciprocal of r is equal to the sum of the reciprocals of x and y" ---> 1/r = (1/x) + (1/y)

" What is r in terms of x and y?" ---> 1 = [(1/x) + (1/y)] * r ----> r = 1 * [(x/1) + (y/1)] ---> r = x + y

Where's the flaw in my calculation?

$$\frac{1}{r} = \frac{1}{x} + \frac{1}{y}$$;

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

$$r=\frac{xy}{x+y}$$.

Hope it's clear.
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Math Expert
Joined: 02 Sep 2009
Posts: 27215
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Kudos [?]: 41011 [0], given: 5654

Re: In an electric circuit, two resistors with resistances x and [#permalink]  29 Dec 2012, 04:42
Expert's post
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

The wording is a bit confusing, though basically we are told that 1/r = 1/x + 1/y, from which it follows that r=xy/(x + y).

_________________
Manager
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Re: In an electric circuit, two resistors with resistances x and [#permalink]  29 Dec 2012, 04:54
From the statements:
x+y = r --> 1
1/r = 1/x+1/y --> 2

From 1 and 2

So 1/r = (x+y)/xy,

r = xy/(x+y)

Ans - D
Manager
Joined: 03 Nov 2009
Posts: 65
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Kudos [?]: 10 [0], given: 17

Re: In an electric circuit, two resistors with resistances x and [#permalink]  29 Dec 2012, 04:54
From the statements:
x+y = r --> 1
1/r = 1/x+1/y --> 2

From 1 and 2

So 1/r = (x+y)/xy,

r = xy/(x+y)

Ans - D
Manager
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Re: In an electric circuit, two resistors with resistances x and [#permalink]  27 Dec 2013, 09:31
Bunuel wrote:
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

The wording is a bit confusing, though basically we are told that 1/r = 1/x + 1/y, from which it follows that r=xy/(x + y).

Could you explain this more in depth?

I interpret it as:

" r is the combined of x and y" --> r = x + y

"the reciprocal of r is equal to the sum of the reciprocals of x and y" ---> 1/r = (1/x) + (1/y)

" What is r in terms of x and y?" ---> 1 = [(1/x) + (1/y)] * r ----> r = 1 * [(x/1) + (y/1)] ---> r = x + y

Where's the flaw in my calculation?
Manager
Joined: 12 Jan 2013
Posts: 249
Followers: 2

Kudos [?]: 32 [0], given: 47

Re: In an electric circuit, two resistors with resistances x and [#permalink]  28 Dec 2013, 04:15
Bunuel wrote:

$$\frac{1}{r} = \frac{1}{x} + \frac{1}{y}$$;

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

$$r=\frac{xy}{x+y}$$.

Hope it's clear.

Bunuel, this is still not clear to me.

I remember asking this very same question (in another thread) about a week ago. Quite obviously there are flaws in my fundamental understanding of concepts in regards to addition of algebraic fractions. Can you direct me to some guide or tutorial that in a simple but efficient way explains these concepts?

Or rather, if you - as you read this - instantly understand where my flaws are and what rule/law/fundamental error I need to "fix", I would be very thankfull if you could explain to me what understanding I lack for algebraic addition of fractions.

By the way, thank you for all of your help!
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In an electric circuit, two resistors with resistances x and [#permalink]  22 Oct 2014, 06:56
aeglorre wrote:
Bunuel wrote:
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

The wording is a bit confusing, though basically we are told that 1/r = 1/x + 1/y, from which it follows that r=xy/(x + y).

Could you explain this more in depth?

I interpret it as:

" r is the combined of x and y" --> r = x + y

"the reciprocal of r is equal to the sum of the reciprocals of x and y" ---> 1/r = (1/x) + (1/y)

" What is r in terms of x and y?" ---> 1 = [(1/x) + (1/y)] * r ----> r = 1 * [(x/1) + (y/1)] ---> r = x + y

Where's the flaw in my calculation?

The steps should be as follows:

-> 1 = [(1/x)+(1/y)] * r
-> r = 1 / [(1/x)+(1/y)]
-> r = 1/ [(y+x)/xy] ... addition in denominator
-> r = 1/ [(x+y)/xy]
-> r = 1/1 * [xy/(x+y)] ... dividing fraction using reciprocal (flipping)
-> r= [xy/(x+y)]
In an electric circuit, two resistors with resistances x and   [#permalink] 22 Oct 2014, 06:56
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