Đáp án:
\(10V\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{R_{AB}} = \dfrac{{\left( {2R + {R_V}} \right)R}}{{3R + {R_V}}} = \dfrac{{2{R^2} + R{R_V}}}{{3R + {R_V}}}\\
{R_{CABD}} = 2R + {R_{AB}} = 2R + \dfrac{{2{R^2} + R{R_V}}}{{3R + {R_V}}} = \dfrac{{8{R^2} + 3R{R_V}}}{{3R + {R_V}}}\\
{R_{CD}} = \dfrac{{{R_V}{R_{CABD}}}}{{{R_V} + {R_{CABD}}}} = \dfrac{{{R_V}.\dfrac{{8{R^2} + 3R{R_V}}}{{3R + {R_V}}}}}{{{R_V} + \dfrac{{8{R^2} + 3R{R_V}}}{{3R + {R_V}}}}} = \dfrac{{8{R^2}{R_V} + 3RR_V^2}}{{8{R^2} + 6R{R_V} + R_V^2}}\\
R = {R_{CD}} + {R_0} = \dfrac{{8{R^2}{R_V} + 3RR_V^2}}{{8{R^2} + 6R{R_V} + R_V^2}} + \dfrac{R}{2} = \dfrac{{22{R^2}{R_V} + 7RR_V^2 + 8{R^3}}}{{2\left( {8{R^2} + 6R{R_V} + R_V^2} \right)}}\\
I = \dfrac{U}{R} = \dfrac{{370\left( {8{R^2} + 6R{R_V} + R_V^2} \right)}}{{22{R^2}{R_V} + 7RR_V^2 + 8{R^3}}}\\
{U_{{R_0}}} = I.{R_0} \Rightarrow 75 = \dfrac{{370\left( {8{R^2} + 6R{R_V} + R_V^2} \right)}}{{22{R^2}{R_V} + 7RR_V^2 + 8{R^3}}}.\dfrac{R}{2}\\
\Rightarrow 75 = \dfrac{{185\left( {8{R^2} + 6R{R_V} + R_V^2} \right)}}{{22R{R_V} + 7R_V^2 + 8{R^2}}}\\
\Rightarrow 880{R^2} - 540R{R_V} - 340R_V^2 = 0\\
\Rightarrow R = {R_V}\\
\Rightarrow I = \dfrac{{150}}{R}\\
{R_{AB}} = \dfrac{3}{4}R \Rightarrow {U_{AB}} = \dfrac{{110}}{{\dfrac{{11}}{4}R}}.\dfrac{3}{4}R = 30V\\
\Rightarrow {U_{{V_2}}} = \dfrac{{30}}{3} = 10V
\end{array}\)
