N cach Cho mach cheh ohei Ru: Ri= Ri= 32 qua mai vāhot 2 den nach tim R di p tA Rs Mar Cnics

Đáp án:

a) \(7,425V\)

b) \(1458J\)

c) \({R_3} = \dfrac{{12}}{7}\Omega \)

Giải thích các bước giải:

a) Ta có: 

\(\begin{array}{l}
{R_{13}} = \dfrac{{{R_1}{R_3}}}{{{R_1} + {R_3}}} = \dfrac{{3.4}}{{3 + 4}} = \dfrac{{12}}{7}\Omega \\
R = {R_{13}} + {R_2} = \dfrac{{12}}{7} + 3 = \dfrac{{33}}{7}\Omega \\
{I_2} = I = \dfrac{E}{{R + r}} = \dfrac{9}{{\dfrac{{33}}{7} + 1}} = 1,575A\\
{I_1} = \dfrac{{{R_3}}}{{{R_1} + {R_3}}}.I = \dfrac{4}{7}.1,575 = 0,9A\\
{I_2} = I - {I_1} = 0,675A\\
U = I.R = 1,575.\dfrac{{33}}{7} = 7,425V
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
{P_1} = I_1^2{R_1} = 0,{9^2}.3 = 2,43W\\
{Q_1} = {P_1}t = 2,43.10.60 = 1458J
\end{array}\)

c) Ta có: 

\(\begin{array}{l}
{R_{13}} = \dfrac{{{R_1}{R_3}}}{{{R_1} + {R_3}}} = \dfrac{{3{R_3}}}{{3 + {R_3}}}\\
R = {R_{13}} + {R_2} = \dfrac{{3{R_3}}}{{3 + {R_3}}} + 3 = \dfrac{{6{R_3} + 9}}{{3 + {R_3}}}\\
{I_2} = I = \dfrac{E}{{R + r}} = \dfrac{9}{{\dfrac{{6{R_3} + 9}}{{3 + {R_3}}} + 1}} = \dfrac{{9\left( {{R_3} + 3} \right)}}{{7{R_3} + 12}}\\
{I_3} = \dfrac{{{R_1}}}{{{R_1} + {R_3}}}.I = \dfrac{3}{{{R_3} + 3}}.\dfrac{{9\left( {{R_3} + 3} \right)}}{{7{R_3} + 12}} = \dfrac{{27}}{{7{R_3} + 12}}\\
{P_3} = I_3^2{R_3} = {\left( {\dfrac{{27}}{{7{R_3} + 12}}} \right)^2}.{R_3} \le \dfrac{{{{27}^2}{R_3}}}{{4.7{R_3}.12}} = \dfrac{{243}}{{112}}{\rm{W}}
\end{array}\)

Dấu = xảy ra khi: \({R_3} = \dfrac{{12}}{7}\Omega \)