Đáp án:

 \({E_2} = 6V\)

Giải thích các bước giải:

* Khi K mở:

Ta có: 

\(\begin{array}{l}
{U_{AB}} = \dfrac{{\dfrac{{{E_1}}}{{{r_1} + {R_1}}} + \dfrac{{{E_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1} + {R_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{R_2}}}}} = \dfrac{{\dfrac{9}{{1 + 2}} + \dfrac{{{E_2}}}{{{r_2}}}}}{{\dfrac{1}{{1 + 2}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{6}}}\\
 \Rightarrow {U_{AB}} = \dfrac{{3 + \dfrac{{{E_2}}}{{{r_2}}}}}{{0,5 + \dfrac{1}{{{r_2}}}}}\\
{I_2} = \dfrac{{{E_2} - {U_{AB}}}}{{{r_2}}} = \dfrac{{{E_2} - \dfrac{{3 + \dfrac{{{E_2}}}{{{r_2}}}}}{{0,5 + \dfrac{1}{{{r_2}}}}}}}{{{r_2}}} = 0\\
 \Rightarrow {E_2} = \dfrac{{3 + \dfrac{{{E_2}}}{{{r_2}}}}}{{0,5 + \dfrac{1}{{{r_2}}}}} \Rightarrow 0,5{E_2} + \dfrac{{{E_2}}}{{{r_2}}} = 3 + \dfrac{{{E_2}}}{{{r_2}}}\\
 \Rightarrow {E_2} = 6V
\end{array}\)

* Khi K đóng:

Ta có: 

\({U_{MN}} = {I_A}.{R_2} = 8,4.6 = 50,4V\)

=> vô lí.