Đáp án:
a) \(160\left( V \right)\) ; \( - 90\left( V \right)\) ; \(35\left( V \right)\) ; \(0\left( V \right)\)
b) \(145\left( V \right)\) ; \(110\left( V \right)\) ; \(20\left( V \right)\)
Giải thích các bước giải:
a) Ta có:
\(\begin{array}{l}
AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{8^2} + {6^2}} = 10\left( {cm} \right)\\
{U_{AC}} = E.AC \Rightarrow 250 = E.0,1 \Rightarrow E = 2500\left( {V/m} \right)\\
{U_{AB}} = E.{d_{AB}} = E.AH = E.AB.\cos A\\
\Rightarrow {U_{AB}} = E.AB.\dfrac{{AB}}{{AC}} = 2500.0,08.\dfrac{8}{{10}} = 160\left( V \right)\\
{U_{CB}} = E.{d_{CB}} = E.\left( { - CH} \right) = - E.BC.\cos C\\
\Rightarrow {U_{CB}} = - 2500.0,06.\dfrac{6}{{10}} = - 90\left( V \right)\\
{U_{MB}} = E.{d_{MB}} = E.MH = E\left( {AH - AM} \right)\\
\Rightarrow {U_{MB}} = 2500\left( {0,08.\dfrac{8}{{10}} - \dfrac{{0,1}}{2}} \right) = 35\left( V \right)\\
{U_{BH}} = E.{d_{BH}} = E.0 = 0\left( V \right)
\end{array}\)
b) Ta có:
\(\begin{array}{l}
{U_{AC}} = 250\left( V \right) \Rightarrow {V_A} - {V_C} = 250\\
\Rightarrow 270 - {V_C} = 250 \Rightarrow {V_C} = 20\left( V \right)\\
{U_{AB}} = 160\left( V \right) \Rightarrow {V_A} - {V_B} = 160\\
\Rightarrow 270 - {V_B} = 160 \Rightarrow {V_B} = 110\left( V \right)\\
{U_{MB}} = 35\left( V \right) \Rightarrow {V_M} - {V_B} = 35\\
\Rightarrow {V_M} - 110 = 35 \Rightarrow {V_M} = 145\left( V \right)\\
{U_{BH}} = 0V \Rightarrow {V_B} = {V_H} = 110\left( V \right)
\end{array}\)
