Short Note
Sum
Solve the following problem.
When two resistors are connected in series, their effective resistance is 80 W. When they are connected in parallel, their effective resistance is 20 W. What are the values of the two resistances? (Answer : 40 W , 40 W)
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Solution
Let the two resistances be `R_1` and `R_2` . Therefore,
`R_1 + R_2 = 80 Ω`
or , `R_1 = 80 - R_2` ....(i)
and
`1/R_1 + 1/R_2 = 1/20`
⇒ `R_1R_2 = 20(R_1+R_2)`
As , `R_1 = 80-R_2`
⇒ `(80 - R_2)R_2 = 20(80 - R_2 + R_2)`
`80R_2^2 - R_2 = 1600`
`R_2^2 - 80R_2 + 1600 = 0`
`(R_2 - 40) (R_2 - 40) = 0`
⇒ `R_2 = 40 Ω`
From (i) , we get
`R_1 = 80-40 =40 Ω`
Concept: Resistance and Ohm'S Law
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