1.

Find the equivalent resistance across the two ends A and B of this circuit.

Answer»

Solution :`(1)/(R_(1,2))=(1)/(R_(1))+(1)/(R_(2))=(1)/(2)+(1)/(2)=(2)/(2)=1 Omega`
SIMILARLY `R_(3,4)= 1 Omega`
`R_(5,6)= 1Omega`
`R_(7,8) = 1 Omega`
Now `R_(1,2) ` and `R_(3,4) ` are connected in SERIES
`R_(1,2,3,4) = R_(1,2)+ R_(3,4)= 1+1=2 Omega`
Also `R_(5,6)` and `R_(7,8)` are connected in series
`R_(5,6,7,8) = R_(5,6)+ R_(7,8) = 1+1= 2 Omega`
Now equivalent resistance
`(1)/(R) = (1)/(R_(1,2,3,4))+(1)/(R_(5,6,7,8))= (1)/(2)+(1)/(2) = (2)/(2) = 1 Omega`
`R =1 Omega`


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