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Question 4  Intext question Pg 216
Asked by tanmaya1.kumar | 28 Mar, 2020, 06:49: PM
answered-by-expert Expert Answer
 
I hope u r asking above question.
 
Let R1 = 2Ω,  R2 = 3Ω  and R3 = 6Ω
 
Then parallel combination of R2 and R3  is given by,  R2||R3 = (3×6)/(3+6) Ω = 2 Ω
 
Hence , if parallel combination of R2 and R3 is connected in series with R1, we get,  R1 + ( R2||R3 ) = 2Ω + 2Ω = 4Ω
 
If all resistors are connected in parallel,  (R1 || R2 || R3 ) = ( 2 × 3 × 6 ) / [ ( 2 × 3 ) + ( 3 × 6 ) + ( 6 × 2 ) ] Ω = (36/36) Ω = 1 Ω 
Answered by Thiyagarajan K | 28 Mar, 2020, 09:01: PM

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