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 1 of 2.
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
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
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
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.
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
VIDYANAND BHARTI said:
1 decade ago
R{EQ)=R1*R2 /R1+R2 =150*6800/150+6800=146.7
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).
Mohammad Shafiq said:
1 decade ago
This is the rule in parallel circuit that the total resistance of parallel circuit is nearly less than from the smallest resistance in parallel circuit so in these answers nearly value is 146.7 and also from formula given above.
Vijay said:
1 decade ago
r1=150,r2=6800
r1*r2=1020000
r1+r2=6950
(r1*r2)/(r1+r2)=1020000/6950
=146.76
r1*r2=1020000
r1+r2=6950
(r1*r2)/(r1+r2)=1020000/6950
=146.76
Dharm singh meena said:
1 decade ago
This rule is sum of the parallel form resistance Rp = r1*r2/r1+r2 150*6800/150+6800 = 147.76.
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