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29 Dec 2012, 05:39
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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
(A) xy
(B) x + y
(C) 1/(x + y)
(D) xy/(x + y)
(E) (x + y)/xy
29 Dec 2012, 05:42
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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
(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).
Answer: D.
28 Dec 2013, 03:42
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aeglorre
Bunuel
Walkabout
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
(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).
Answer: D.
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.
Thanks
