Bài 2:

b,

Vẽ ảnh $A'$, $B'$

Vẽ đường thẳng đi qua hai điểm $A'$; $B'$

Tại đưuòng thẳng ấy ta thấy ảnh của $A$; $B$ chồng lên nhau

Bài 3:

$P_{1}=20W$

$P_{2}=40W$

$U$

$U_{1}$

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Ta có: $P_{1}=\dfrac{U²}{R_{1}}$

$P_{2}=\dfrac{U²}{R_{2}}$

$⇒\dfrac{P_{1}}{P_{2}}=\dfrac{R_{2}}{R_{1}}=\dfrac{1}{2}$

$⇒R_{1}=2.R_{2}$

Ta có: $I_{2}=I_{1}=\dfrac{U}{R_{1}}=\dfrac{U}{2.R_{2}}$

$P_{2}=I_{2}².R_{2}=\dfrac{U²}{4.R_{2}}=\dfrac{40}{4}=10W$

Bài 4:

a, $Sđmđ:$ $[(R_{1}//R_{2})ntR_{3}]//R_{4}$

$R_{12}=\dfrac{R_{1}.R_{2}}{R_{1}+R_{2}}=\dfrac{16.16}{16+16}=8$

$R_{123}=R_{12}+R_{3}=8+4=12$

$R=\dfrac{R_{123}.R_{4}}{R_{123}+R_{4}}=\dfrac{12.12}{12+12}=6$

$I=\dfrac{U}{R}=\dfrac{12}{6}=2A$

$I_{3}=\dfrac{U}{R_{123}}=\dfrac{12}{12}=1A$

$⇒I_{1}+I_{2}=1A$

Ta có: $I_{1}.R_{1}=I_{2}.R_{2}$

Mà $R_{1}=R_{2}$

$⇒I_{1}=I_{2}=0,5A$

$⇒I_{A}=I-I_{1}=2-0,5=1,5$

b, $Sđmđ:$ $[(R_{2}ntR_{4})//R_{3})ntR_{1}$

$R_{24}=R_{2}+R_{4}=16+12=28$

$R_{243}=\dfrac{R_{24}.R_{3}}{R_{24}+R_{3}}=\dfrac{28.4}{28+4}=3,5$

$R=R_{243}+R_{1}=3,5+16=19,5$

$I=\dfrac{U}{R}=\dfrac{12}{19,5}=\dfrac{8}{13}A$

Ta có: $\dfrac{U_{1}}{R_{1}}=\dfrac{U_{3}}{R_{243}}$

$⇔\dfrac{U_{1}}{16}=\dfrac{12-U_{1}}{3,5}$

$⇔U_{1}=\dfrac{128}{13}V$

$⇒U_{2}+U_{4}=U_{3}=12-U_{1}=\dfrac{28}{13}$

Ta có: $\dfrac{U_{2}}{R_{2}}=\dfrac{U_{4}}{R_{4}}$

$⇔\dfrac{\dfrac{28}{13}-U_{4}}{16}=\dfrac{U_{4}}{12}$

$⇔U_{4}=\dfrac{12}{13}V$

$U_{v}=U-U_{4}=12-\dfrac{12}{13}=\dfrac{144}{13}V$