tion:Solution :-**refer to attached for the reference.First of all we will add some more information in the diagram in ORDER to solve it.As In AMG , A battery is connected we will take a current i_1i 1 Then ACCORDING to Kirchoff's junction law at junction G.i_1 = I + xi 1 =I+xx = i_1 - Ix=i 1 −ISo current in GB = i_1 - Ii 1 −INow in FED we will take another current i_2i 2 .I = i_2 + yI=i 2 +yy = I - i_2y=I−i 2 So current in FC = I - i_2I−i 2 Now by APPLYING junction law at other junctions▪️Current in CB = II▪️Current in BA = i_1i 1 Now we will write Loop For (anticlockwise)(1)AHGB-12 = 3R(i_1 - I) + R(i_1)−12=3R(i 1 −I)+R(i 1 )\implies -12 = 3Ri_1 - 3RI + Ri_1⟹−12=3Ri 1 −3RI+Ri 1 \implies -12 = 4Ri_1 - 3RI⟹−12=4Ri 1 −3RI ...(i)(2) AHFC-12 = 4R(I - i_2) + R(i_1)−12=4R(I−i 2 )+R(i 1 )\implies -12 = 4RI - 4Ri_2 + Ri_1⟹−12=4RI−4Ri 2 +Ri 1 ...(ii)(3) AHED24 - 12 = 2R(i_2) + R(i_1)24−12=2R(i 2 )+R(i 1 )\implies12 = 2Ri_2 + Ri_1⟹12=2Ri 2 +Ri 1 ....(iii)(4)CFED24 = 2R(i_2) - 4R(I-i_2)24=2R(i 2 )−4R(I−i 2 )\implies 24 = 2Ri_2 - 4RI + 4Ri_2⟹24=2Ri 2 −4RI+4Ri 2 \implies 24 = 6Ri_2 - 4RI⟹24=6Ri 2 −4RI ....(iv)Now 3(iv) - 4(i)72 = 18Ri_2 - 12RI72=18Ri 2 −12RI+ 48 = - 16Ri_1 + 12RI+48=−16Ri 1 +12RI________________________120 = 18Ri_2 - 16Ri_1120=18Ri 2 −16Ri 1 .....(v)________________________Now from (v) - 9(iii)120 = 18Ri_2 - 16Ri_1120=18Ri 2 −16Ri 1 -108= - 18Ri_2 - 9Ri_1−108=−18Ri 2 −9Ri 1 ________________________12 = -25Ri_112=−25Ri 1 ________________________Ri_1 = \dfrac{-12}{25}Ri 1 = 25−12 Now from by substiting value of Ri_1Ri 1 in (i)-12 = 4Ri_1 - 3RI−12=4Ri 1 −3RI\implies -12 = 4\dfrac{-12}{25} - 3RI⟹−12=4 25−12 −3RI\implies -12 = \dfrac{-48}{25}- 3RI⟹−12= 25−48 −3RI\implies 3RI = \dfrac{-48}{25} + 12⟹3RI= 25−48 +12\implies 3RI = \dfrac{300-48}{25}⟹3RI= 25300−48 \implies 3RI = \dfrac{252}{25}⟹3RI= 25252 \implies 3RI = 10.08⟹3RI=10.08\implies RI = 3.36⟹RI=3.36\implies I = \dfrac{3.36}{R}⟹I= R3.36 \implies I = \dfrac{3.36}{1000}⟹I= 10003.36 \implies I = 3.36 \: mA⟹I=3.36mASo current I = 3.36 mAfollow me