CBSE Class 10 Answered
finding the net resistance
![question image](https://images.topperlearning.com/topper/new-ate/6413113651ad9eb78f84Screenshot20231002202133.png)
Asked by daniya062008 | 02 Oct, 2023, 20:25: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/aa7042cf1035d615ca7d6b77d08dddba651ae6cb8f21c0.29327541f8.png)
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](https://images.topperlearning.com/topper/tinymce/cache/706b46d00af38324b483833480cc5110.png)
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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](https://images.topperlearning.com/topper/tinymce/cache/83b59efac0fd92fde42725adb4428dcf.png)
Answered by Thiyagarajan K | 02 Oct, 2023, 21:45: PM
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