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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. - Science and Technology

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`

  Is there an error in this question or solution?
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APPEARS IN

Balbharati Science and Technology 9th Standard Maharashtra State Board
Chapter 3 Current Electricity
Exercise | Q 9.2 | Page 45
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