Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 2)
2.
What is the product-over-sum result of 150 and 6800?
Discussion:
17 comments Page 2 of 2.
Ravikumar said:
1 decade ago
Name suggested product over sum means first product the given values and then sum it now results of both are divided prodect over sum example: two values say R1&R2 the resultant product of sum is R= (R1*R2) / (R1+R2).
VIDYANAND BHARTI said:
1 decade ago
R{EQ)=R1*R2 /R1+R2 =150*6800/150+6800=146.7
Vikas said:
1 decade ago
This is law of parallel addition of resistance.
R = r1*r2/r1+r2 = 146.7
R = r1*r2/r1+r2 = 146.7
S. Mohammed Audhil said:
1 decade ago
Friends,
Parallel resistance R=(R1*R2)/(R1+R2)
where R1, R2 are the two parallel resistors
Hence,
R = (150*6800)/(150+6800) = 146.7.
Parallel resistance R=(R1*R2)/(R1+R2)
where R1, R2 are the two parallel resistors
Hence,
R = (150*6800)/(150+6800) = 146.7.
Thara said:
1 decade ago
sum=150+6800
product=150*6800
therefore
product over sum=1020000/6950
=146.7
product=150*6800
therefore
product over sum=1020000/6950
=146.7
Laran said:
1 decade ago
product=150*6800=1020000
sum=150+6800=6950
product-over-sum=1020000/6950=146.7
sum=150+6800=6950
product-over-sum=1020000/6950=146.7
Purwa upadhyay said:
1 decade ago
Formula of: product-over-sum r=(r1*r2)/(r1+r2)
r=(150*6800)/(150+6800)
r=146.7ohm
r=(150*6800)/(150+6800)
r=146.7ohm
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