Explanation:
Applying Kirchhoff's first law at the junction \(P\), we get
\(6 = {I_1} + {I_2}\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Applying Kirchhoff's secod law to the closed circuit PQRP, we get
\( - 2I - 2{I_1} + 2{I_2} = 0\)
\(or,\,\,{I_2} = 2{I_1}\,\,\,\,\,\,\,\,\,(2)\)
Solve (1) and (2), we get
\({I_1} = 2A,{I_2} = 4A\)