Đáp án:
\(\begin{array}{l}
a.R = \dfrac{{88}}{7}\Omega \\
b.\\
U = \dfrac{{176}}{7}V\\
{U_V} = \dfrac{{120}}{7}V\\
c.\\
{P_1} = 16W\\
{P_2} = \dfrac{{720}}{{49}}{\rm{W}}\\
{P_3} = \dfrac{{960}}{{49}}{\rm{W}}\\
d.Q = \dfrac{{63360}}{7}J = 2162,413cal
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{20.15}}{{20 + 15}} = \dfrac{{60}}{7}\Omega \\
R = {R_1} + {R_{23}} = 4 + \dfrac{{60}}{7} = \dfrac{{88}}{7}\Omega \\
b.\\
U = {\rm{IR}} = 2.\dfrac{{88}}{7} = \frac{{176}}{7}V\\
{U_V} = {U_{23}} = {\rm{I}}{{\rm{R}}_{23}} = 2.\dfrac{{60}}{7} = \dfrac{{120}}{7}V\\
c.\\
{P_1} = {R_1}{I_1}^2 = {4.2^2} = 16W\\
{U_2} = {U_3} = {U_{23}} = \dfrac{{120}}{7}V\\
{P_2} = \dfrac{{U_2^2}}{{{R_2}}} = \dfrac{{{{\dfrac{{120}}{7}}^2}}}{{20}} = \dfrac{{720}}{{49}}{\rm{W}}\\
{P_3} = \dfrac{{U_3^2}}{{{R_3}}} = \dfrac{{{{\dfrac{{120}}{7}}^2}}}{{15}} = \dfrac{{960}}{{49}}{\rm{W}}\\
d.\\
Q = UIt = \dfrac{{176}}{7}.2.180 = \dfrac{{63360}}{7}J = 2162,413cal
\end{array}\)
