Đáp án:
\(\begin{array}{l}
a.\\
{Q_1} = {6.10^{ - 4}}\\
{Q_2} = 1,{5.10^{ - 3}}\\
420V = {U_1} = {U_2}\\
W = 0,441\\
b.\\
{Q_1}' = 3,{6.10^{ - 4}}\\
{Q_2}' = 5,{4.10^{ - 4}}\\
{U_1} = {U_2} = 180V\\
W = 0,081
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{Q_1} = {C_1}{U_1} = {2.10^{ - 6}}.300 = {6.10^{ - 4}}\\
{Q_2} = {C_2}{U_2} = {3.10^{ - 6}}500 = 1,{5.10^{ - 3}}\\
{C_b} = {C_1} + {C_2} = {2.10^{ - 6}} + {3.10^{ - 6}} = {5.10^{ - 6}}\\
{Q_b} = {Q_1} + {Q_2} = {6.10^{ - 4}} + 1,{5.10^{ - 3}} = 2,{1.10^{ - 3}}\\
U = \frac{{{Q_b}}}{{{C_b}}} = \frac{{2,{{1.10}^{ - 3}}}}{{{{5.10}^{ - 6}}}} = 420V = {U_1} = {U_2}\\
W = \frac{1}{2}{C_b}{U^2} = \frac{1}{2}{.5.10^{ - 6}}.420 = 0,441\\
b.\\
{Q_1} - {Q_2} = - {Q_1}' - {Q_2}'\\
{Q_1}' + {Q_2}' = - {6.10^{ - 4}} + 1,{5.10^{ - 3}} = {9.10^{ - 4}} = {Q_b}\\
{U_1}' = {U_2}'\\
\frac{{{Q_1}'}}{{{C_1}}} = \frac{{{Q_2}'}}{{{C_2}}} \Leftrightarrow \frac{{{Q_1}'}}{2} = \frac{{{Q_2}'}}{3}\\
{Q_1}' = 3,{6.10^{ - 4}}\\
{Q_2}' = 5,{4.10^{ - 4}}\\
{C_b} = {C_1} + {C_2} = {2.10^{ - 6}} + {3.10^{ - 6}} = {5.10^{ - 6}}\\
U = \frac{{{Q_b}}}{{{C_b}}} = \frac{{{{9.10}^{ - 4}}}}{{{{5.10}^{ - 6}}}} = 180V\\
W = \frac{1}{2}{C_b}{U^2} = \frac{1}{2}{5.10^{ - 6}}{.180^2} = 0,081
\end{array}\)
