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

Options

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

Answer

C

Explanation

0.13 A, from Q to P

By applying Krichhoff.s `2^(nd)` rule on loop FPDCQAF
`- 2I + 6 - 3 (I - I_(1)) - 3I = -9`
`therefore - 2I - 3I + 3I_(1) - 3I = - 9 - 6 `
`therefore 8 I - 3I_(1) = 15 " "`...(I)
By applying Kirchhoff.s `2^(nd)` rule on loop FPQAF,
` -2I - I_(1) - 3I = - 9 `
` therefore 5I + I_(1) = 9 " "` .... (2)
By multiplying equation (1) with (5) and equation (2) with (8) and then subtracting ,
`underset("-23 "I_(1) =3)overset overset overset ( "40I "-15I_(1) = 75)("40I "+8I_(1) = 72)(-" "-" "-)(_)`
`therefore I_(1) = - (3)/(23)`
`therefore I_(1) = 0.13 `A
Value of current is negative so that current is flowing from Q to P.