in a science experiment,when resistor A and resister B are connected in a parallel circuit,the total resistance is 1 all over 1/A + 1/B.This complex fraction is equivalent to
2007-11-16 10:21:57 · 7 answers · asked by yanksrule4691 1 in Science & Mathematics ➔ Mathematics
[02]
1/A +1/B
=(A+B)/AB
Therefore 1 all over 1/A +1/B
=AB/(A+B)
2007-11-16 10:28:53 · answer #1 · answered by alpha 7 · 0⤊ 0⤋
1/2
2007-11-16 10:26:18 · answer #2 · answered by Anonymous · 0⤊ 1⤋
AB/B+A
because you start from the bottom of the fraction and work your way to the top.
so, you need a common denominator to add 1/A to 1/B. so multiply 1/A by B(top and bottom) and 1/B by A so that you get ... B/AB + A/AB ... then you can add them together making ... B+A/AB ... then, you wanna eliminate the bottom fraction so you have only one top fraction. so you multiply B+A/AB by the reverse (AB/B+A) so you can cross multiply and cancel that one out. but, to make it even, you have to multiply the top number(1) by the same thing so 1 x AB/B+A which equals AB/B+A...... ta da!
2007-11-16 10:29:31 · answer #3 · answered by Anonymous · 0⤊ 0⤋
2ab.
1 times each or
2(a+b)
2007-11-16 10:25:48 · answer #4 · answered by S81 5 · 0⤊ 1⤋
1 / (1/A + 1/B) = 1 / [ (B/AB) + A/(AB) ] = 1 / [ (A + B) / AB ]
---> = AB / (A + B)
2007-11-16 10:28:03 · answer #5 · answered by Aeons 2 · 0⤊ 0⤋
[02]
1/A +1/B
=(A+B)/AB
So,1 all over 1/A +1/B
=AB/(A+B)
2007-11-16 10:32:49 · answer #6 · answered by Anonymous · 0⤊ 0⤋
AB/(A+B)
2007-11-16 10:27:40 · answer #7 · answered by ironduke8159 7 · 0⤊ 0⤋