Find the resistance between point A and B in the circuit diagram given below:
Text Solution
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`R_(1)=6Omega` `R_(2),R_(3),R_(4)` are in series and have resultant resistance R' `R'=R_(2)+R_(3)+R_(4)=2+2+2=6Omega` R' is in parallel combination with `R_(1)` `therefore` Resultant resistance of the circuit (R ) `rArr (1)/(R)=(1)/(R_(1))+(1)/(R_(1))=(1)/(6)+(1)/(6)=(2)/(6)=(1)/(3) " " therefore "Resistance",R=3Omega`
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