Cho mạch điện (Hình 2): các điện trở R1...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

1. Ta có:  \(\left\{ \begin{array}{l}{I_A} = \frac{U}{{{R_1} + {R_D}}} = \frac{U}{{4R}} = 1A\\P = \frac{{3{U^2}}}{{4R}} = 36W\end{array} \right. \Rightarrow \left\{ \begin{array}{l}R = 3\Omega \\U = 12V\end{array} \right.\)

2a. Ta có :

\({P_2} = 8W \Rightarrow \left\{ \begin{array}{l}{I_2} = \sqrt {\frac{{{P_2}}}{{{R_2}}}}  = \sqrt {\frac{8}{3}} A\\{U_{CB}} = \sqrt {{P_2}{R_2}}  = 2\sqrt 6 V\end{array} \right. \Rightarrow {I_1} = \frac{{{U_{AC}}}}{{{R_1}}} = \frac{{U - {U_{CB}}}}{{{R_1}}} = 4 - \sqrt {\frac{8}{3}} V\)

Ta có :

\(\begin{array}{l}\left\{ \begin{array}{l}{I_X} = {I_1} - {I_2} = {I_3} - {I_D}\\U = {R_D}{I_D} + {R_3}{I_3}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{I_3} - {I_D} = 4 - \sqrt {\frac{{32}}{3}} \\4 = 3{I_D} + {I_3}\end{array} \right.\\{I_D} = \sqrt {\frac{2}{3}} A \Rightarrow {P_D} = {R_D}.I_D^2 = 6W\end{array}\)

2b.

Ta có: \(\left\{ \begin{array}{l}{I_2} = {I_1} - {I_X}\\{U_{AB}} = U = {R_1}{I_1} + {R_2}{I_2}\end{array} \right. \Rightarrow U = R(2{I_1} - {I_X})\,\,\,\,(1)\)

Và: \(\left\{ \begin{array}{l}{I_3} = {I_X} + {I_D}\\U = {R_D}{I_D} + {R_3}{I_3}\end{array} \right. \Rightarrow U = R(4{I_D} + {I_X})\,\,\,\,\,\,(2)\)

\(\buildrel {(1)(2)} \over
\longrightarrow {I_1} = 2{I_D} + {I_X}\,\,\,\,\,\,\,\,\,\,(3)\)

 Lại có:

\(\eqalign{
& {U_{AD}} = {R_D}{I_D} = {R_1}{I_1} + {R_X}{I_X}\,\,\,\,\,\,\,\,\,\,\,\,\,(4) \cr
& \buildrel {(3);(4)} \over
\longrightarrow R{I_D} = (R + {R_X}){I_X}\,\,\,\,\,\,\,\,\,\,\,\,\,(5) \cr
& \buildrel {(2);(5)} \over
\longrightarrow {I_X} = {{\rm{U}} \over {5R + 4{R_X}}} \Rightarrow {P_X} = {R_X}I_X^2 = {{{{\rm{U}}^2}{R_X}} \over {{{\left( {5R + 4{R_X}} \right)}^2}}} \Rightarrow {P_{\max }} = {{{U^2}} \over {80R}} = 0,6W \Leftrightarrow {R_X} = {5 \over 4}R = 3,75\Omega \cr} \)

 2c. \({I_D} = \frac{U}{{R\left( {4 + \frac{R}{{R + {R_X}}}} \right)}}\) => R_X tăng  => I_D tăng hay độ sáng tăng.