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Bài 49:

* Khi K mở, mạch gồm: (R1 nt R2) // (Rv1 nt Rv2)

Ta có:

\({R_{{V_1}}} = {R_{{V_2}}} \Rightarrow {U_{{V_1}}} = {U_{{V_2}}} = \dfrac{U}{2} = \dfrac{{120}}{2} = 60V\)

* Khi K đóng, mạch gồm: (R1 // Rv1) nt (R2 // Rv2)

Ta có:

\(\begin{array}{l}
{R_{1V}} = \dfrac{{{R_1}{R_V}}}{{{R_1} + {R_V}}} = \dfrac{{1000.4000}}{{1000 + 4000}} = 800\Omega \\
{R_{2V}} = \dfrac{{{R_2}{R_V}}}{{{R_2} + {R_V}}} = \dfrac{{2000.4000}}{{2000 + 4000}} = \dfrac{{4000}}{3}\Omega \\
I = \dfrac{U}{{{R_{1V}} + {R_{2V}}}} = \dfrac{{120}}{{800 + \dfrac{{4000}}{3}}} = \dfrac{9}{{160}}A\\
{U_{{V_1}}} = I.{R_{1V}} = \dfrac{9}{{160}}.800 = 45V\\
{U_{{V_2}}} = U - {U_{{V_1}}} = 120 - 45 = 75V
\end{array}\)

Bài 50:

* K mở, mạch gồm: R0 nt 2R

Ta có:

\({U_1} = \dfrac{U}{{{R_0} + 2R}}.{R_0} = 6\)

* K đóng, mạch gồm: R0 nt (2R // R)

Ta có:

\(\begin{array}{l}
{R_{td}} = {R_0} + \dfrac{{2R.R}}{{2R + R}} = {R_0} + \dfrac{2}{3}R = \dfrac{{3{R_0} + 2R}}{3}\\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{3U}}{{3{R_0} + 2R}}\\
{U_2} = I.{R_0} = \dfrac{{3U{R_0}}}{{3{R_0} + 2R}} = 10\\
 \Rightarrow \dfrac{{\dfrac{{3U{R_0}}}{{3{R_0} + 2R}}}}{{\dfrac{U}{{{R_0} + 2R}}.{R_0}}} = \dfrac{{10}}{6}\\
 \Rightarrow \dfrac{{3\left( {{R_0} + 2R} \right)}}{{3{R_0} + 2R}} = \dfrac{{10}}{6}\\
 \Rightarrow 18\left( {{R_0} + 2R} \right) = 10\left( {3{R_0} + 2R} \right)\\
 \Rightarrow 16R = 12{R_0}\\
 \Rightarrow \dfrac{R}{{{R_0}}} = \dfrac{3}{4}
\end{array}\)

HĐT giữa 2 đầu mạch là:

\({U_{BD}} = \dfrac{{{U_1}}}{{{R_0}}}\left( {{R_0} + 2R} \right) = {U_1} + 2{U_1}.\dfrac{R}{{{R_0}}} = 15V\)