Đáp án:
\(\begin{array}{l}
a.\\
{R_0} = \frac{{225}}{{19}}\Omega \\
b.\\
{U_d} = U(1 - \frac{1}{{1 + \frac{{{R_{1d}}}}{{{R_0}}}}})\\
c.\\
{P_{\max }} = \frac{{304}}{{15}}W\\
{R_0} = \frac{{135}}{{19}}\Omega
\end{array}\)
b.
khi con chạy C qua phải thì R0 giảm thì Ud tăng nên đàng sáng mạnh hơn
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_d} = \frac{{U_{dm}^2}}{{{P_{dm}}}} = \frac{{{9^2}}}{6} = 13,5\Omega \\
{I_d} = \frac{{{P_{dm}}}}{{{U_{dm}}}} = \frac{6}{9} = \frac{2}{3}A\\
{U_0} = U - {U_d} = 24 - 9 = 15V\\
{I_1} = \frac{{{U_d}}}{{{R_1}}} = \frac{9}{{15}} = 0,6A\\
I = {I_d} + {I_1} = 0,6 + \frac{2}{3} = \frac{{19}}{{15}}A\\
{R_0} = \frac{U}{I} = \frac{{15}}{{\frac{{19}}{{15}}}} = \frac{{225}}{{19}}\Omega \\
b.\\
{U_d} = U - {U_0} = U - I{R_0} = U - \frac{U}{R}{R_0} = U(1 - \frac{{{R_0}}}{{{R_0} + {R_{1d}}}}) = U(1 - \frac{1}{{1 + \frac{{{R_{1d}}}}{{{R_0}}}}})\\
c.\\
{R_{1d}} = \frac{{{R_1}{R_d}}}{{{R_1} + {R_d}}} = \frac{{15.13,5}}{{15 + 13,5}} = \frac{{135}}{{19}}\Omega \\
P = {R_0}{I^2} = {R_0}{(\frac{U}{R})^2} = {R_0}\frac{{{U^2}}}{{{{({R_0} + {R_{1d}})}^2}}} = \frac{{{U^2}}}{{{{(\sqrt {{R_0}} + \frac{{{R_{1d}}}}{{{R_0}}})}^2}}}\\
{P_{\max }} \Rightarrow \sqrt {{R_0}} + \frac{{{R_{1d}}}}{{{R_0}}}\min \\
\sqrt {{R_0}} + \frac{{{R_{1d}}}}{{{R_0}}} \ge 2\sqrt {{R_{1d}}} \\
\Rightarrow {P_{\max }} = \frac{{{U^2}}}{{4{R_{d1}}}} = \frac{{{{24}^2}}}{{4.\frac{{135}}{{19}}}} = \frac{{304}}{{15}}W\\
{R_0} = {R_{1d}} = \frac{{135}}{{19}}\Omega
\end{array}\)
