In the given circuit, the current in the circuit is

Options

A. `0.45 A`
B. `0.27 A`
C. `0.1 A`
D. `0.37 A`

Answer

A

Explanation

(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