12 câu trắc nghiệm Vật lí 11 Kết nối tri thức Bài 23: Điện trở. Định luật Ohm (Đúng sai - Trả lời ngắn) có đáp án

Hướng dẫn giải:

a. $S = \pi \frac{{{d^2}}}{4} = \pi \frac{{{{({{1.10}^{ - 3}})}^2}}}{4} \approx 7,{85.10^{ - 7}}({m^2})$

b. $R = \rho \frac{l}{S} = \frac{{1,{{69.10}^{ - 8}}.2}}{{\pi \frac{{{{({{1.10}^{ - 3}})}^2}}}{4}}} \approx 0{,}043(\Omega )$

c. $U = I.R = 1{,}5.0{,}043 \approx 0{,}065(V)$

d. $U = I'.R = 3.0{,}043 \approx 0{,}13(V)$

Hướng dẫn giải:

a. U$_{3}$=I$_{3}$.R$_{3}$=2.10=20V

b. I$_{5}$=I$_{3}$=I$_{35}$=2A

R$_{35}$=10+20=30 Ω

$$\begin{array}{l}{U_{35}} = {U_2} = {U_{235}} = {I_{35}}.{R_{35}} = 2.30 = 60V\\{I_2} = \frac{{{U_2}}}{{{R_2}}} = \frac{{60}}{{20}} = 3A\end{array}$$

c.

$$\begin{array}{l}{R_{235}} = \frac{{{U_{235}}}}{{{I_{235}}}} = \frac{{60}}{{3 + 2}} = 12\Omega \\{R_{2345}} = {R_4} + {R_{235}} = 20 + 12 = 32\Omega \\R = \frac{{32.8}}{{32 + 8}} = 6{,}4\Omega \end{array}$$

d.

$$\begin{array}{l}{U_{35}} = {U_2} = {U_{235}} = {I_{35}}.{R_{35}} = 2.30 = 60V\\{I_2} = \frac{{{U_2}}}{{{R_2}}} = \frac{{60}}{{20}} = 3A\end{array}$$

Hướng dẫn giải:

a. R = 100+200=300 Ω

b. $I = \frac{{12}}{{300}} = 0{,}04(A)$

c. ${U_1} = I.{R_1} = 0{,}04.100 = 4V$

d. ${U_2} = I.{R_2} = 0{,}04.200 = 8V$

a. $R = \frac{{{6^2}}}{3} = 12\Omega$

b. ${I_{dm}} = \frac{{{P_{dm}}}}{{{U_{dm}}}} = \frac{3}{6} = 0{,}5A$

c. $\begin{array}{l}{U_2} = {U_D} = 6V\\{U_1} = U - {U_{2D}} = 9 - 6 = 3V\\{I_1} = {I_{2D}} = \frac{{{U_1}}}{{{R_1}}} = \frac{3}{2} = 1{,}5A\\{R_{2D}} = \frac{{{U_{2D}}}}{{{I_{2D}}}} = \frac{6}{{1{,}5}} = 4\Omega \\\frac{{12.{R_2}}}{{12 + {R_2}}} = 4 \Rightarrow {R_2} = 6\Omega \end{array}$

d. I$_{2}$=I$_{1}$-I$_{đm}$=1,5-0,5=1A

Hướng dẫn giải:

***a.* **$R = 5 + 10 + 3 = 18\Omega$

b. ${I_1} = {I_2} = {I_3} = I = \frac{U}{R} = \frac{{18}}{{18}} = 1A$

***c.* **${U_1} = {I_1}{R_1} = 1.5 = 5V$

***d.* **${U_2} = {I_2}{R_2} = 1.10 = 10V$

Hướng dẫn giải:

***a.* **${R_{D1}} = \frac{{U_1^2}}{{{P_1}}} = \frac{{2,{5^2}}}{1} = 6{,}25\Omega$

***b.* **${R_{D2}} = \frac{{U_2^2}}{{{P_2}}} = \frac{{{6^2}}}{3} = 12\Omega$

***c.* **

$R = \frac{{({R_1} + {R_{D1}}).{R_{D2}}}}{{({R_1} + {R_{D1}}) + {R_{D2}}}} = \frac{{(8{,}75 + 6{,}25)12}}{{(8{,}75 + 6{,}25) + 12}} = \frac{{20}}{3}\Omega$

***d.* **

$\begin{array}{l}{I_{D1}} = {I_1} = {I_{dm1}} = \frac{1}{{2{,}5}} = 0{,}4(A)\\{I_{D2}} = {I_{dm2}} = \frac{3}{6} = 0{,}5(A)\\I = {I_{D1}} + {I_{D2}} = 0{,}4 + 0{,}5 = 0{,}9(A)\end{array}$

Đáp án:

3

,

6

$\begin{array}{l}{R_{12}} = {R_1} + {R_2} = 6 + 3 = 9\Omega \\R = \frac{{{R_{12}}.{R_3}}}{{{R_{12}} + {R_3}}} = \frac{{9.6}}{{9 + 6}} = 3{,}6\Omega \end{array}$

Đáp án:

8

$\begin{array}{l}U = {U_{12}} = 12(V)\\{I_{12}} = {I_1} = {I_2} = \frac{{{U_{12}}}}{{{R_{12}}}} = \frac{{12}}{9} = \frac{4}{3}(A)\\{U_1} = {I_1}.{R_1} = \frac{4}{3}6 = 8(V)\end{array}$

Đáp án:

1

,

9

6

$S = \frac{{\rho l}}{R} = \frac{{1,{{1.10}^{ - 6}}0{,}8}}{{4{,}5}} \approx 1,{96.10^{ - 7}}({m^2})$

Đáp án:

5

$$S = \pi \frac{{{d^2}}}{4} \Leftrightarrow 1,{96.10^{ - 7}} = \pi \frac{{{d^2}}}{4} \Rightarrow d \approx {5.10^{ - 4}}(m)$$

Đáp án:

5

6

,

2

$m = V.D = D.S.l \Leftrightarrow 0{,}5 = {8900.1.10^{ - 6}}.l \Rightarrow l \approx 56{,}2(m)$

Đáp án:

0,

9

6

Hướng dẫn giải

$R = \rho \frac{l}{S} = 1,{7.10^{ - 8}}.\frac{{56{,}2}}{{10{}^{ - 6}}} \approx 0{,}96\Omega$