In the circuit shown, the current in the `1Omega` resistor is :

Options

A. `0.13 A, from Q to P `
B. `0.13 A, from P to Q`
C. `1.3 A from P to Q`
D. 0A

Answer

A

Explanation

(a) From KVL
`-6 + 3I_1 + 1(I_i-I_2) = 0`

`6 = 3I_1 + I_1-I_2`
`4I_1-I_2 = 6` ……..(1)
`-9+2I_2 - (I_1 -I_2)+3I-2=0`
`-I_1 + 6I_2 = 9` ………..(2)
On solving (1) and (2)
`I_1 = 0.13A`
Direction Q to P, since`I_1gtI_2`.