Ta có mạch R3nt(R1//R2)
Đặt R2=x
Ta có \(Rt\text{đ}=R3+\dfrac{R1.R2}{R1+R2}=4+\dfrac{6.x}{6+x}=\dfrac{24+10x}{6+x}\)
\(I=\dfrac{U}{Rt\text{đ}}=6:\dfrac{24+10x}{6+x}=\dfrac{36+6x}{24+10x}\)
vì R3ntR12=> \(I3=I12=I=\dfrac{36+6x}{24+10x}\)
Ta có U1=I1.R1=\(\dfrac{1}{3}.6=2V\)
Vì R1//R2=> U1=U2=U12=I12.R12=\(\dfrac{36+6x}{24+10x}.\dfrac{6.x}{6+x}=2=>x=3\Omega\)
=> R2=3\(\Omega\)