Đoạn mạch có năm điện trở R1 và R2,R3, R4, R5 mắc nối tiếp. Biết R1=2R2=3R3=4R4=5R5.Hiệu điện thế giưa hai đầu đoạn mạch là 50V. Tính hiệu điện thế giữa hai đầu mỗi điện trở?

lengochien13022006

2 năm trước

Loga Sinh Học lớp 12

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quadrat99

Đáp án:

$U_1=\dfrac{3000}{137}V$

$U_2=\dfrac{1500}{137}V$

$U_3=\dfrac{1000}{137}V$

$U_4=\dfrac{750}{137}V$

$U_5=\dfrac{600}{137}V$

Giải thích các bước giải:

$R_1=2R_2=3R_3=4R_4=5R_5$

$⇒U_1=2U_2=3U_3=4U_4=5U_5$

$R_1ntR_2ntR_3ntR_4ntR_5$

$R_{tđ}=R_1+R_2+R_3+R_4+R_5$

$⇒ R_{tđ}=R_1+\dfrac{R_1}{2}+\dfrac{R_1}{3}+\dfrac{R_1}{4}+\dfrac{R_1}{5}=\dfrac{137}{60}R_1$

Ta có: $I_1=I_2=I_3=I_4=I_5=I$

$I_1=I$

$⇒\dfrac{U_1}{R_1}=\dfrac{U}{R_{tđ}}$

$⇒\dfrac{U_1}{R_1}=\dfrac{50}{\dfrac{137}{60}R_1}$

$⇒U_1=\dfrac{3000}{137}V$

$⇒U_2=\dfrac{1500}{137}V$

$⇒U_3=\dfrac{1000}{137}V$

$⇒U_4=\dfrac{750}{137}V$

$⇒U_5=\dfrac{600}{137}V$

Vote (0) Phản hồi (0) 2 năm trước

82tinh

Đáp án:

$\begin{array}{l}
{U_1} = \dfrac{{3000}}{{137}}V\\
{U_2} = \dfrac{{1500}}{{137}}V\\
{U_3} = \dfrac{{1000}}{{137}}V\\
{U_4} = \dfrac{{750}}{{137}}V\\
{U_5} = \dfrac{{600}}{{137}}V
\end{array}$

Giải thích các bước giải:

Hiệu điện thế giữa 2 đầu mỗi điện trở là:
$\begin{array}{l}
{I_1} = {I_2} = {I_3} = {I_4} = {I_5} = I\\
\Leftrightarrow \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{{{U_4}}}{{{R_4}}} = \dfrac{{{U_5}}}{{{R_5}}} = \dfrac{U}{{{R_1} + {R_2} + {R_3} + {R_4} + {R_5}}}\\
\Leftrightarrow \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{{U_2}}}{{\dfrac{{{R_1}}}{2}}} = \dfrac{{{U_3}}}{{\dfrac{{{R_1}}}{3}}} = \dfrac{{{U_4}}}{{\dfrac{{{R_1}}}{4}}} = \dfrac{{{U_5}}}{{\dfrac{{{R_1}}}{5}}} = \dfrac{U}{{{R_1} + \dfrac{{{R_1}}}{2} + \dfrac{{{R_1}}}{3} + \dfrac{{{R_1}}}{4} + \dfrac{{{R_1}}}{5}}}\\
\Leftrightarrow {U_1} = \dfrac{U}{{1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5}}} = \dfrac{{50}}{{\dfrac{{137}}{{60}}}} = \dfrac{{3000}}{{137}}V\\
\Rightarrow {U_2} = \dfrac{{{U_1}}}{2} = \dfrac{{3000}}{{137.2}} = \dfrac{{1500}}{{137}}V\\
\Rightarrow {U_3} = \dfrac{{{U_1}}}{3} = \dfrac{{3000}}{{3.137}} = \dfrac{{1000}}{{137}}V\\
\Rightarrow {U_4} = \dfrac{{{U_1}}}{4} = \dfrac{{3000}}{{4.137}} = \dfrac{{750}}{{137}}V\\
\Rightarrow {U_5} = \dfrac{{{U_1}}}{5} = \dfrac{{3000}}{{5.137}} = \dfrac{{600}}{{137}}V
\end{array}$

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