a, khi Khóa K mở =>\([\left(R1ntR4\right)//R2]ntR3\)
=>\(R1234=\dfrac{\left(R1+R4\right).R2}{R1+R4+R2}+R3=\dfrac{\left(40+R4\right)90}{130+R4}+40\left(om\right)\)
=>\(I1234=\dfrac{350}{R1234}=\dfrac{350}{\dfrac{\left(40+R4\right)90}{130+R4}+40}\left(A\right)=I124\)
có \(U124=I124.R124\)
\(< =>\left(U1+U4\right)=\dfrac{350}{\dfrac{\left(40+R4\right)90}{130+R4}+40}.\dfrac{\left(40+R4\right)90}{130+R4}\)
\(< =>\left(I4.R1+I4.R4\right)=\dfrac{350}{\dfrac{\left(40+R4\right)90}{130+R4}+40}.\dfrac{\left(40+R4\right)90}{130+R4}\)
\(< =>\left(90+2,25R4\right)=\dfrac{350}{\dfrac{\left(40+R4\right)90}{130+R4}+40}.\dfrac{\left(40+R4\right)90}{130+R4}\)
giải pt trên=>R4=40(ôm)
b, K đóng vẽ lại mạch R1//[R2nt(R3//R4)]
\(R_{234}=20+90=110\left(\Omega\right)\)
\(I_{234}=\dfrac{350}{110}=\dfrac{35}{11}\left(A\right)\)
\(\Rightarrow U_4=U_{34}=\dfrac{35}{11}.20\approx63,63\left(V\right)\)
