Please send q 15?

Apply the #Y-Delta# transformation to the #3# resistors on the #RHS#

That is #R_1=50 Omega#, #R_3=15 Omega# and #R_2=20 Omega#

#R_a=(R_1R_2)/(R_1+R_2+R_3)=(50*20)/(50+15+20)=1000/85=11.76 Omega#

#R_b=(R_2R_3)/(R_1+R_2+R_3)=(20*15)/(4+15+10)=300/85=3.53Omega#

#R_c=(R_3R_1)/(R_1+R_2+R_3)=(50*15)/(4+15+10)=750/85=8.82Omega#

#i_1+i_2=1.4A#

and

#U=(R_c+4)i_2=(R_b+10)i_2#

Therefore,

#(8.82+10)(1.4-i_1)=(7.53)*i_2#

Solving for #i_1#

#18.82*1.4-18.82i_2=7.53i_2#

#i_2=26.35/26.35=1.0A#

Ratio of #"R"_1# and #"R"_2# is

#"R"_1/"R"_2 = "20 Ω"/"50 Ω" = 2/5#

Ratio of #"R"_3# and #"R"_4# is

#"R"_3/"R"_4 = "4 Ω"/"10 Ω" = 2/5#

As the ratio is same in both the cases it’s a balanced Wheatstone bridge. Therefore, #"15 Ω"# resistance can be neglected.

Circuit now:

In series, resistances are added. So the circuit now is

Let #"i"_1# be the current passing through #"24 Ω"# and #"i"_2# be the current passing through #"60 Ω"#. Then,

#"Current" ∝ 1/"Resistance" color(white)(...)[∵ "V = constant"]#

#"i"_1/"i"_2 = "60 Ω"/"24 Ω" = 5/2#

#"i"_1 = 5/(5 + 2) × "1.4 A" = "1.0 A"#

#"i"_2 = 2/(5 + 2) × "1.4 A" = "0.4 A"#

∴ Current passing through #"4 Ω"# and #"20 Ω"# is #"1.0 A"#.