Câu 1.
\(I_A=I_m=1A\)\(\Rightarrow I_R=1A\Rightarrow U_R=I_R\cdot R=1\cdot4=4V\)
\(\Rightarrow U_V=U_{xV}=U-U_R=12-4=8V\)
\(R_{xV}=\dfrac{U_{xV}}{I_{xV}}=\dfrac{8}{1}=8\Omega\)
Mà \(\dfrac{1}{R_{xV}}=\dfrac{1}{R_x}+\dfrac{1}{R_V}=\dfrac{1}{R_x}+\dfrac{1}{2000}=\dfrac{1}{8}\)
\(\Rightarrow R_x=\dfrac{2000}{249}\approx8,03\Omega\)
