Câu 18:

a) Ta có: 

\(\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{n_1}}}{{{n_2}}} \Rightarrow \dfrac{{800}}{{{U_2}}} = \dfrac{{600}}{{12000}} \Rightarrow {U_2} = 16000V\)

b) Công suất hao phí là:

\({P_{hp}} = {\left( {\dfrac{P}{{{U_2}}}} \right)^2}R = {\left( {\dfrac{{{{10.10}^3}}}{{{{16.10}^3}}}} \right)^2}.40 = 15,625W\)

Câu 19: 

a) Đây là ảnh thật.

b) Ta có: 

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{30}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{{20}}{{OA' - 20}}\\
 \Rightarrow \dfrac{{30}}{{OA'}} = \dfrac{{20}}{{OA' - 20}} \Rightarrow OA' = 60cm
\end{array}\)

Câu 20: 

a) Ta có: \({N_2} > {N_1} \Rightarrow \) đây là máy tăng thế.

b) Ta có: 

\(\dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{n_1}}}{{{n_2}}} \Rightarrow \dfrac{{400}}{{{U_2}}} = \dfrac{{500}}{{40000}} \Rightarrow {U_2} = 32000V\)

c) Công suất hao phí là:

\({P_{hp}} = {\left( {\dfrac{P}{{{U_2}}}} \right)^2}R = {\left( {\dfrac{{{{10}^6}}}{{32000}}} \right)^2}.40 = 39062,5W\)

d) Để công suất hao phí giảm 1 nửa thì: 

\({U_2}' = {U_2}\sqrt 2  = 32000\sqrt 2 V\)

Câu 21: 

Ta có:

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} \Rightarrow \dfrac{2}{4} = \dfrac{{10}}{{OA'}} \Rightarrow OA' = 20cm\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{f}{{20 - f}}\\
 \Rightarrow \dfrac{1}{2} = \dfrac{f}{{20 - f}} \Rightarrow f = \dfrac{{20}}{3}cm
\end{array}\)