1.

In the given circuit, the current through 2 Omega resistor is

Answer»

`0.45 A`
`0.27 A`
`0.1 A`
`0.37 A`

Solution :(a): `R_(3)` and `R_(4)` are in series. So, `R' = R_(3) + R_(4) = 1+1=2OMEGA`

Now, `R_(2),R_(5)` and R' are in parallel, So,
`1/(R’’) = 1/(R_(2)) + 1/(R_(5)) + 1/(R’) = 1/1 + 1/1 + 1/2 = 5/2 rArr R’’ = 2/5 Omega`
Now, `R_(7), R_(8)` and R’’ are in parallel, So
`1/(R’’’) = 1/3 + 1/1 +5/2 rArr` R’’’ `6/(23) Omega`
Now, `R_(1),R_(8)` and R’’’ are in series.
So, EQUIVALENT resistance, `R=1 + 6/(23) + 2 = (75)/(23) Omega`
`:.` Current through `2Omega` resistor is,
`I = V/R = (1.2xx23)/(75) = 0.36` A


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