Trắc nghiệm Điện trở. Định luật Ohm lớp 11 (có đúng sai, trả lời ngắn)

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. U₃=I₃.R₃=2.10=20V

b. I₅=I₃=I₃₅=2A

R₃₅=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. I₂=I₁-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.

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