Electricity And Magnitism, Resistance

The net resistance of two resistors in parallel combination is 2Ω and in series combination is 9Ω. The two resistors are

6Ω , 3Ω 6Ω , 9Ω 2Ω , 6Ω 6Ω , 2Ω

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1 solution

Efren Pineda
Aug 21, 2014

Resistors in Parallel be equation (1): R 1 R 2 R 1 + R 2 = 2 ( 1 ) \frac{R_1R_2}{R_1+R_2}=2~~~~(1) Resistors in Series be equation (2): R 1 + R 2 = 9 ( 2 ) R_1+R_2=9~~~~(2) Plug-in (2) to (1) yields R 1 R 2 9 = 2 \frac{R_1R_2}{9}=2 R 1 R 2 = 18 ( 3 ) R_1R_2=18~~~~(3) From (2)... R 2 = 9 R 1 R_2=9 - R_1 , plug-in at (3) will yield R 1 ( 9 R 1 ) = 18 R_1(9 - R_1)=18 9 R 1 R 1 2 = 18 9R_1 - R_1^2=18 Re-arranging terms and equate to zero to solve R 1 R_1 , R 1 2 9 R 1 + 18 = 0 R_1^2 - 9R_1+18=0 By Factoring, we have ( R 1 3 ) ( R 1 6 ) = 0 (R_1-3)(R_1-6)=0 therefore if R 1 = 3 Ω R_1=3\Omega , R 2 = 6 Ω R_2=6 \Omega and

if R 1 = 6 Ω R_1=6\Omega , R 2 = 3 Ω R_2=3 \Omega .

Why are you doing it by such a complex method. once u have got the product of the 2 resistances just take out it's factors as 6x3, and as6+3=9. therefore the answer is 6&3ohms respectively.

Sourabh shukla - 6 years, 6 months ago

Or we can just look at the options and find out which two add up to 9

Sutirtha Paul - 5 years, 3 months ago

How do you know that 6 Ω , and 3 Ω 6\Omega, \text { and } 3\Omega can be only?

. . - 1 month, 3 weeks ago

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