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finding the net resistance  
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Asked by daniya062008 | 02 Oct, 2023, 08:25: PM
answered-by-expert Expert Answer
Case (a)
 
Resistors of resitances 1Ω and 2Ω are connected in series . This series combination is connected in parallel to 1.5Ω resistor.
 
Equivalent resistance of series combination of 1Ω and 2Ω is 3Ω .
 
Then equivalent resistance of 3Ω resistor  connected parallel to 1.5 Ω  is
 
begin mathsize 14px style R subscript e q end subscript space equals space fraction numerator 3 cross times 1.5 over denominator 3 plus 1.5 end fraction space capital omega space equals space 1 space capital omega end style
 
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Case (b)
 
Resistors of resistances 3 Ω and 2 Ω above diagonal line AB are in series . Equivalent resistance of this series combination is 5Ω.
 
Similarly , Resistors of resistances 3 Ω and 2 Ω below diagonal line AB are in series . Equivalent resistance of this series combination is 5Ω.
 
Hence 5 Ω resistance above diagonal AB,  5 Ω resistance below diaonal AB and 1 Ω resistance in the diagonal line AB all are connected in parallel.
 
Equivalent resistance of network given in case (b) is
 
begin mathsize 14px style R subscript e q end subscript space equals space fraction numerator 5 cross times 1 cross times 5 over denominator 5 cross times 1 space plus space 1 cross times 5 space plus space 5 space cross times 5 end fraction space equals 0.714 space capital omega end style
Answered by Thiyagarajan K | 02 Oct, 2023, 09:45: PM

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