1.

Find the equivalent resistance across the 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`
`:. R_(1,2 )=1 OMEGA`
Similarly `R_(3,4) =1 Omega`
`R_(5,6) =1 Omega`
`R_(7,8) = 1Omega`
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 = 2Omega`
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 = 2Omega`
Now, equivalent RESISTANCE
`1/R_(eq) = 1/(R_(1,2,3,4)) + 1/(R_(5,6,7,8))`
`=1/2 + 1/2=2 /2 =1 `
`:. R_(eq) = 1Omega`


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