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draw arrengement R1 R2 R4 R4 so that net resistance of the resultant circuit is (i)26 ohm (ii) minimum R1=10Ω, R2= 5Ω, R3= 15Ω, R4= 20Ω
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Asked by wierdtechs | 27 Jul, 2018, 02:01: PM
answered-by-expert Expert Answer
5Ω and 15Ω in series with parallel combination of 20Ω and 10Ω gives the equivalent reasistance 26.6 Ω
 
5+15+(20||10) = 20+(200/30) = 26.6 Ω
 
If all resistances are connected in parallel, then we get minimum resiatane 3.4 Ω
Answered by Thiyagarajan K | 27 Jul, 2018, 03:19: PM

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