a, k mở vẽ lại mạch ((R1ntR4)//R2)ntR3
\(I_1=I_4=2\left(A\right)\)
\(I_2+2=I_3\)
\(U_{AB}=\left(I_2+2\right)R_3+I_2R_2\Rightarrow I_2=\dfrac{5}{3}\left(A\right)\)
\(U_4=U_2-U_1=I_2R_2-I_1R_1=80\left(V\right)\Rightarrow R_4=\dfrac{80}{2}=40\left(\Omega\right)\)
b, k đóng mạch vẽ lại R1//[R2nt(R3//R4)]
\(\Rightarrow I_{234}=\dfrac{420}{R_{234}}=\dfrac{420}{144}=\dfrac{35}{12}\left(A\right)\)
\(\Rightarrow U_4=U_{34}=I_{234}.R_{34}=\dfrac{35}{12}.24=70\left(V\right)\)
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