Câu hỏi của Song Hậu Nguyễn

Question

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nguyễn thị hương giang · Accepted Answer

$R_{13}=\dfrac{4\cdot2}{4+2}=\dfrac{4}{3}\Omega$

$R_N=\dfrac{\dfrac{4}{3}\cdot6}{\dfrac{4}{3}+6}=\dfrac{12}{11}\Omega$

$U_{AB}=12+8=20V$

$I=\dfrac{U}{R}=\dfrac{20}{\dfrac{12}{11}}=\dfrac{55}{3}A$

$I_{13}=I_m=\dfrac{55}{3}A\Rightarrow U_{13}=\dfrac{55}{3}\cdot\dfrac{4}{3}=\dfrac{220}{9}V$

$\Rightarrow U_3=\dfrac{220}{9}V\Rightarrow I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{220}{9}}{2}=\dfrac{110}{9}A$

Câu trả lời:

$R_{13}=\dfrac{4\cdot2}{4+2}=\dfrac{4}{3}\Omega$

$R_N=\dfrac{\dfrac{4}{3}\cdot6}{\dfrac{4}{3}+6}=\dfrac{12}{11}\Omega$

$U_{AB}=12+8=20V$

$I=\dfrac{U}{R}=\dfrac{20}{\dfrac{12}{11}}=\dfrac{55}{3}A$

$I_{13}=I_m=\dfrac{55}{3}A\Rightarrow U_{13}=\dfrac{55}{3}\cdot\dfrac{4}{3}=\dfrac{220}{9}V$

$\Rightarrow U_3=\dfrac{220}{9}V\Rightarrow I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{220}{9}}{2}=\dfrac{110}{9}A$