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Asked by Sunita | 24 Sep, 2018, 09:18: PM
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Given resistance network is redrawn in step-1.
 
we see that across points C and D, 9Ω resistor is in parallel with a series combination of 10Ω and 8Ω resistor.
It can be worked out that across C and D equivalent resistance is 6Ω.
Similarly across G and H, we have a 20Ω resistor parallel to a series combination of resistors 15Ω and 5Ω.
Equivalent restsance between the points G and H can be calculated as 10Ω. 

Step-2 is shown in figure after doing the above mentioned simplification.
Now across E and F, we have 48Ω is parallel to series combination of 6Ω and 10Ω.
Equivalent resistance between E and F can be calculated as 12Ω.

Step-3 in figure shows the final simplification.
Equivalent resistance of three resistors 14Ω, 6Ω and 12Ω which are in series is 32Ω
Answered by Thiyagarajan K | 25 Sep, 2018, 06:23: AM

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