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

(b) Applying Kirchhoff's law for the loop (1) and (2) as shown in figure
For loop (1)
`- 2 i_(1) - 2 (i_(1) - i_(2)) + 4 = 0 implies 2 i_(1) + i_(2) = 2`
For loop (2)
`- 2 (i_(1) - i_(2)) + 4 i_(2) - 6 = 0 implies - i_(1) + 3 i_(2) = 3`
On solving equations (i) and (ii) `i_(1) = 1.8 A`