In the circuit shown below `E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega` and `R_(3) = 2 Omega`. The current `I_(1)` is

Options

A. 1.6 A
B. 1.8 A
C. 1.25 A
D. 1.0 A

Answer

B

Explanation

`4-2i_(1) + 2(i_(2) - i_(1))=0`
`rArr 2i_(2) - 4i_(1)+ = 0`………. (i)
`6-(i_(2)-i_(1))2 - 4i_(2) = 0`
`rArr -6i_(2) + 2i_(1) + 6=0`……….(ii)
From equations (i) and (ii)
`6i_(2) - 12i_(1) + 12=0`
`-6i_(2) + 2i_(1) + 6=0`
`10i_(1) = 18 rArr i_(1) = 18/10 = 1.8 A`