A current of A enters one corner P of an equilateral triangle PQR having 3 wires of resistances `2 Omega` each and leaves by the corner R. Then the current `I_(1)` and `I_(2)` are

Options

A. 2 A, 4 A
B. 4 A, 2 A
C. 1 A, 2 A
D. 2 A, 3 A

Answer

A

Explanation

Applying Kirchhoff's first law at the junction P, we get
`6=I_(1)+I_(2) " "` …..(i)
Applying Kirchhoff's second law to the closed circuit PQRP, we get
`-2I_(1)-2I_(1)+2I_(2)=0`
or, `2I_(1)+2I_(1)-2I_(2)=0`
or, `4I_(1)-2I_(2)=0 " "` ....(ii)
Solve (i) and (ii), we get
`I_(1)=2 A, I_(2)=4 A`.