Đáp án:

 $\begin{align}
  & a){{U}_{2}}=7,127V \\ 
 & b){{R}_{V}}=900\Omega  \\ 
 & {{U}_{AB}}=19V \\ 
\end{align}$

Giải thích các bước giải:

 ${{R}_{1}}=300\Omega ;{{R}_{2}}=225\Omega ;{{U}_{V1}}=9,5V;$

Sơ đồ mạch điện : $V//{{R}_{1}}$

ta có: Điện trở toàn mạch:

$\begin{align}
  & {{R}_{AB}}=\dfrac{{{R}_{1}}.{{R}_{V}}}{{{R}_{1}}+{{R}_{V}}}+{{R}_{2}} \\ 
 & =\dfrac{{{R}_{1}}.{{R}_{V}}+{{R}_{1}}{{R}_{2}}+{{R}_{2}}{{R}_{V}}}{{{R}_{1}}+{{R}_{V}}} \\ 
\end{align}$

mà: $\frac{{{U}_{1}}}{{{U}_{AB}}}=\dfrac{{{R}_{1}}{{V}_{1}}}{{{R}_{1}}{{R}_{2}}+{{R}_{V}}.({{R}_{1}}+{{R}_{2}})(1)}$

Sơ đồ mạch điện : $V//{{R}_{2}}$

Tương tự ta có: 

$\begin{array}{*{35}{l}}
   \text{ }\!\!~\!\!\text{ } & R{{'}_{AB}}={{R}_{1}}+\dfrac{{{R}_{2}}.{{R}_{V}}}{{{R}_{2}}+{{R}_{V}}}  \\
   \text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{ } & =\dfrac{{{R}_{1}}.{{R}_{V}}+{{R}_{1}}{{R}_{2}}+{{R}_{2}}{{R}_{V}}}{{{R}_{2}}+{{R}_{V}}}  \\
   \text{ }\!\!~\!\!\text{ } & {}  \\
\end{array}$

$\dfrac{{{U}_{2}}}{U{{'}_{AB}}}=\dfrac{{{R}_{2}}V}{{{R}_{1}}{{R}_{2}}+{{R}_{V}}.({{R}_{1}}+{{R}_{2}})}(2)$

Từ (1) và (2) ta có:

$\begin{align}
  & \dfrac{{{U}_{1}}}{{{U}_{2}}}=\dfrac{{{R}_{1}}}{{{R}_{2}}} \\ 
 & \Rightarrow {{U}_{2}}={{U}_{1}}.\dfrac{{{R}_{2}}}{{{R}_{1}}}=9,5.\dfrac{225}{300}=7,125V \\ 
\end{align}$

b)$Vnt{{R}_{1}}nt{{R}_{2}}$

Từ (1) ta có:

$\begin{align}
  & \frac{{{U}_{1}}}{{{U}_{AB}}}=\dfrac{{{R}_{1}}{{R}_{V}}}{{{R}_{1}}{{R}_{2}}+{{R}_{V}}.({{R}_{1}}+{{R}_{2}})} \\ 
 & \Rightarrow {{U}_{1}}=\dfrac{{{R}_{1}}{{R}_{V}}}{{{R}_{1}}{{R}_{2}}+{{R}_{V}}.({{R}_{1}}+{{R}_{2}})}.{{U}_{AB}}(3) \\ 
\end{align}$

Khi nối tiếp ta có:

$\begin{align}
  & R={{R}_{1}}+{{R}_{2}}+{{R}_{V}} \\ 
 & \Rightarrow I=\dfrac{{{U}_{AB}}}{R}=\dfrac{{{U}_{AB}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{V}}} \\ 
 & \dfrac{{{U}_{V}}}{{{U}_{AB}}}=\frac{{{R}_{V}}}{R}=\dfrac{{{R}_{V}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{V}}} \\ 
 & \Rightarrow {{U}_{V}}=\dfrac{{{R}_{V}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{V}}}.{{U}_{AB}}=12V(4) \\ 
\end{align}$

Từ (3) và (4):

$\begin{align}
  & \dfrac{{{U}_{1}}}{{{U}_{V}}}=\dfrac{{{R}_{1}}({{R}_{1}}+{{R}_{V}})+{{R}_{1}}{{R}_{V}}}{{{R}_{1}}{{R}_{2}}+{{R}_{V}}.({{R}_{1}}+{{R}_{2}})} \\ 
 & \Leftrightarrow \dfrac{9,5}{12}=\frac{R_{1}^{2}+{{R}_{1}}{{R}_{V}}+{{R}_{1}}{{R}_{V}}}{{{R}_{1}}{{R}_{2}}+{{R}_{1}}{{R}_{V}}+{{R}_{2}}{{R}_{V}}} \\ 
 & \Rightarrow {{R}_{V}}=\dfrac{12R_{1}^{2}+2,5{{R}_{1}}{{R}_{2}}}{9,5{{R}_{2}}-2,5{{R}_{1}}}=\dfrac{{{12.300}^{2}}+2,5.300.225}{9,5.225-2,5.300}=900\Omega  \\ 
\end{align}$

Hiệu điện thế:

${{U}_{AB}}=12.\dfrac{300+225+900}{900}=19V$