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13 tháng 3 2018
bài 1
ta có \(\Delta ABO\) \(\sim\Delta A'B'O\)(g.g)
\(\Rightarrow\dfrac{AB}{A'B'}=\dfrac{OA}{OA'}\)
\(\Leftrightarrow\dfrac{2,5}{1}=\dfrac{20}{OA'}\)
\(\Rightarrow OA'=\dfrac{20.1}{2,5}=8\left(cm\right)\)
5 tháng 10 2021
\(a,\)\(R1.R4=R2.R3\Rightarrow\dfrac{R1}{R2}=\dfrac{R3}{R4}\Rightarrow\left(R1ntR3\right)//\left(R2ntR4\right)\)
\(\Rightarrow\left\{{}\begin{matrix}R1.R4=20^2=400\left(\Omega\right)\Rightarrow R1=\dfrac{400}{R4}\left(1\right)\\I13=\dfrac{U}{R13}=\dfrac{18}{R1+R3}=\dfrac{18}{R1+20}\left(A\right)\\I24=\dfrac{18}{R2+R4}=\dfrac{18}{R4+20}\left(A\right)\end{matrix}\right.\)
\(\Rightarrow Im=I13+I24=\dfrac{18}{R1+20}+\dfrac{18}{R4+20}=\dfrac{18}{Rtd}=\dfrac{18}{\dfrac{\left(R1+R3\right)\left(R2+R4\right)}{R1+R2+R3+R4}}=\dfrac{18}{\dfrac{\left(20+R1\right)\left(20+R4\right)}{R1+R4+40}}\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\dfrac{18}{\dfrac{400}{R4}+20}+\dfrac{18}{R4+20}=\dfrac{18}{\dfrac{\left(\dfrac{400}{R4}+20\right)\left(R4+20\right)}{\dfrac{400}{R4}+R4+40}}\Rightarrow\left\{{}\begin{matrix}R4=5\Omega\\R1=\dfrac{400}{5}=80\Omega\end{matrix}\right.\)
\(\Rightarrow Rtd=\dfrac{\left(R1+R3\right)\left(R2+R4\right)}{R1+R2+R3+R4}=\dfrac{\left(20+80\right)\left(20+5\right)}{20+80+20+5}=20\Omega\)
\(b,\Rightarrow\left(R3//R2\right)nt\left(R1//R4\right)\Rightarrow\)\(Ia=0,3A=I3-I1\)
\(\Rightarrow\dfrac{I4}{I1}=\dfrac{R1}{R4}\Rightarrow I1=\dfrac{R4.I4}{R1}=\dfrac{R4\left(Im-I1\right)}{R1}\left(A\right)\)
\(\Rightarrow Im=\dfrac{18}{Rtd}=\dfrac{18}{\dfrac{R2R3}{R2+R3}+\dfrac{R1R4}{R1+R4}}=\dfrac{18}{10+\dfrac{400}{R1+R4}}\left(A\right)\)
\(\Rightarrow I1=\dfrac{U-U23}{R1}=\dfrac{18-Im.R23}{R1}=\dfrac{18-\dfrac{180}{10+\dfrac{400}{R1+R4}}}{R1}=\dfrac{\dfrac{180+\dfrac{7200}{R1+R4}-180}{10+\dfrac{400}{R1+R4}}}{R1}=\dfrac{\dfrac{\dfrac{7200}{R1+R4}}{\dfrac{10R1+10R4+400}{R1+R4}}}{R1}=\dfrac{\dfrac{7200}{10R1+10R4+400}}{R1}=\dfrac{7200}{R1\left(10R1+10R4+400\right)}=\dfrac{7200}{10R1^2+400R1+4000}\left(A\right)\)
\(\Rightarrow I3+I2=Im\Rightarrow I3=\dfrac{Im}{2}=\dfrac{\dfrac{18}{10+\dfrac{400}{R1+R4}}}{2}=\dfrac{9}{10+\dfrac{400}{R1+R4}}\left(A\right)\)
\(\Rightarrow\dfrac{9}{10+\dfrac{400}{R1+R4}}-\dfrac{7200}{10R1^2+400R1+4000}=0,3\Rightarrow\dfrac{9}{10+\dfrac{400}{R1+\dfrac{400}{R1}}}-\dfrac{7200}{10R1^2+400R1+4000}=0,3\Rightarrow\left\{{}\begin{matrix}R1=40\Omega\\R4=10\Omega\end{matrix}\right.\)\(\)
\(\Rightarrow Rtd=\dfrac{Rac.Rd}{Rac+Rd}+\dfrac{Rbc.R1}{Rbc+R1}=\dfrac{12.\left(\dfrac{6^2}{6}\right)}{12+\dfrac{6^2}{6}}+\dfrac{12.12}{12+12}=10\Omega\)
\(b,\Rightarrow Ibc1=\dfrac{U}{Rtd}=\dfrac{12}{10}=1,2A\Rightarrow Ubc1=Ibc1\left(\dfrac{Rbc.R1}{Rbc+R1}\right)=7,2V\Rightarrow I1=\dfrac{7,2}{R1}=0,6A\Rightarrow Q1=I1^2R1t=1296W\)
\(c,\Rightarrow\left\{{}\begin{matrix}Ud=6V=Uac\\Id=\dfrac{Pdm}{Udm}=1A\end{matrix}\right.\)
\(\Rightarrow\)\(\dfrac{12}{Rtd}=\dfrac{12}{\dfrac{Rac.Rd}{Rac+Rd}+\dfrac{\left(24-Rac\right)R1}{24-Rac+R1}}=\dfrac{12}{\dfrac{6Rac}{6+rac}+\dfrac{\left(24-Rac\right).12}{36-Rac}}=Iacd\)
\(\Rightarrow1+Iac=Iacd\Rightarrow1+\dfrac{6}{Rac}=\dfrac{12}{\dfrac{6Rac}{6+Rac}+\dfrac{\left(24-Rac\right)12}{36-Rac}}\Rightarrow Rac=12\sqrt{2}\left(\Omega\right)\)
3. R4 nt {R1//(R2ntR3)}
\(a,\Leftrightarrow\)\(Ia=0,3A=I2=I3\Rightarrow U23=U123=I2.\left(R2+R3\right)=6V\)
\(\Rightarrow Im=\dfrac{U123}{R123}=\dfrac{6}{\dfrac{R1\left(R2+R3\right)}{R1+R2+R3}}=0,5A\Rightarrow Uab=Im.Rtd=0,5\left(R4+R123\right)=10V\)
\(b,\) R2//{R1 nt(R3//R4)}
\(\Rightarrow K\) mở \(\Rightarrow I3=\dfrac{U.R123}{Rtd.R23}=\dfrac{6}{12+R4}\left(A\right)\)
\(\Rightarrow K\) đóng \(\Rightarrow I3=\dfrac{U.R4}{R134.R34}=\dfrac{2R4}{30+7R4}\left(A\right)\)
\(\Rightarrow R4=15\Omega\)
\(\Rightarrow Ik=I2+I3=\dfrac{U}{R2}+\dfrac{2.15}{30+7.15}=\dfrac{10}{15}+\dfrac{2.15}{30+7.15}=\dfrac{8}{9}A\)