Three resistors R1- 1 - - R2 - 100 - - R3 - 1000 - are connected in parallel- Which relation holds true-

1/Req. = 1/R_1 + 1/R_2 +1/R_3 1/Req.= nbsp; nbsp;1/1 + 1/100 +1/1000 1/ Req = nbsp; 1 + 0.01 + 0.0001 nbsp; 1/ Req = 1.0101 nbsp;Req = 1/ 1.0101 Req ...

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Standard X

Physics

Combination of Resistances

Three resisto...

Question

Three resistors R1= 1 Ω , R2 = 100 Ω , R3 = 1000 Ω are connected in parallel. Which relation holds true?

Req. < 1

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1 > Req> 100

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Req = 100

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1 > Req > 1000

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Solution

The correct option is A Req. < 1
$\frac{1}{R e q .}$ = $\frac{1}{R_{1}}$ + $\frac{1}{R_{2}}$ + $\frac{1}{R_{3}}$
1/Req.= $\frac{1}{1}$ + $\frac{1}{100}$ + $\frac{1}{1000}$
1/ Req = 1 + 0.01 + 0.0001
1/ Req = 1.0101
Req = 1/ 1.0101
Req. < 1 ( 0.98 )
Req. < 1

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Three resistors R1= 1 Ω , R2 = 100 Ω , R3 = 1000 Ω are connected in parallel. Which relation holds true?

The correct option is A Req. < 11Req. = 1R1 + 1R2 +1R3 1/Req.= 11 + 1100 +11000 1/ Req = 1 + 0.01 + 0.0001 1/ Req = 1.0101 Req = 1/ 1.0101 Req. < 1 ( 0.98 ) Req. < 1

The correct option is A Req. < 1
$\frac{1}{R e q .}$ = $\frac{1}{R_{1}}$ + $\frac{1}{R_{2}}$ + $\frac{1}{R_{3}}$
1/Req.= $\frac{1}{1}$ + $\frac{1}{100}$ + $\frac{1}{1000}$
1/ Req = 1 + 0.01 + 0.0001
1/ Req = 1.0101
Req = 1/ 1.0101
Req. < 1 ( 0.98 )
Req. < 1