K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
1 tháng 10 2021
\(\Rightarrow Rtd=\dfrac{Rac.Rd}{Rac+Rd}+\dfrac{Rbc.R1}{Rbc+R1}=\dfrac{12.\left(\dfrac{6^2}{6}\right)}{12+\dfrac{6^2}{6}}+\dfrac{12.12}{12+12}=10\Omega\)
\(b,\Rightarrow Ibc1=\dfrac{U}{Rtd}=\dfrac{12}{10}=1,2A\Rightarrow Ubc1=Ibc1\left(\dfrac{Rbc.R1}{Rbc+R1}\right)=7,2V\Rightarrow I1=\dfrac{7,2}{R1}=0,6A\Rightarrow Q1=I1^2R1t=1296W\)
\(c,\Rightarrow\left\{{}\begin{matrix}Ud=6V=Uac\\Id=\dfrac{Pdm}{Udm}=1A\end{matrix}\right.\)
\(\Rightarrow\)\(\dfrac{12}{Rtd}=\dfrac{12}{\dfrac{Rac.Rd}{Rac+Rd}+\dfrac{\left(24-Rac\right)R1}{24-Rac+R1}}=\dfrac{12}{\dfrac{6Rac}{6+rac}+\dfrac{\left(24-Rac\right).12}{36-Rac}}=Iacd\)
\(\Rightarrow1+Iac=Iacd\Rightarrow1+\dfrac{6}{Rac}=\dfrac{12}{\dfrac{6Rac}{6+Rac}+\dfrac{\left(24-Rac\right)12}{36-Rac}}\Rightarrow Rac=12\sqrt{2}\left(\Omega\right)\)
1 tháng 10 2021
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 \(\Rightarrow I3=\dfrac{U.R4}{R134.R34}=\dfrac{2R4}{30+7R4}\left(A\right)\)
\(\Rightarrow R4=15\Omega\)
\(\Rightarrow Ik=I2+I3=\dfrac{U}{R2}+\dfrac{2.15}{30+7.15}=\dfrac{10}{15}+\dfrac{2.15}{30+7.15}=\dfrac{8}{9}A\)
5 tháng 10 2021
\(a,\)\(R1.R4=R2.R3\Rightarrow\dfrac{R1}{R2}=\dfrac{R3}{R4}\Rightarrow\left(R1ntR3\right)//\left(R2ntR4\right)\)
\(\Rightarrow\left\{{}\begin{matrix}R1.R4=20^2=400\left(\Omega\right)\Rightarrow R1=\dfrac{400}{R4}\left(1\right)\\I13=\dfrac{U}{R13}=\dfrac{18}{R1+R3}=\dfrac{18}{R1+20}\left(A\right)\\I24=\dfrac{18}{R2+R4}=\dfrac{18}{R4+20}\left(A\right)\end{matrix}\right.\)
\(\Rightarrow Im=I13+I24=\dfrac{18}{R1+20}+\dfrac{18}{R4+20}=\dfrac{18}{Rtd}=\dfrac{18}{\dfrac{\left(R1+R3\right)\left(R2+R4\right)}{R1+R2+R3+R4}}=\dfrac{18}{\dfrac{\left(20+R1\right)\left(20+R4\right)}{R1+R4+40}}\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\dfrac{18}{\dfrac{400}{R4}+20}+\dfrac{18}{R4+20}=\dfrac{18}{\dfrac{\left(\dfrac{400}{R4}+20\right)\left(R4+20\right)}{\dfrac{400}{R4}+R4+40}}\Rightarrow\left\{{}\begin{matrix}R4=5\Omega\\R1=\dfrac{400}{5}=80\Omega\end{matrix}\right.\)
\(\Rightarrow Rtd=\dfrac{\left(R1+R3\right)\left(R2+R4\right)}{R1+R2+R3+R4}=\dfrac{\left(20+80\right)\left(20+5\right)}{20+80+20+5}=20\Omega\)
\(b,\Rightarrow\left(R3//R2\right)nt\left(R1//R4\right)\Rightarrow\)\(Ia=0,3A=I3-I1\)
\(\Rightarrow\dfrac{I4}{I1}=\dfrac{R1}{R4}\Rightarrow I1=\dfrac{R4.I4}{R1}=\dfrac{R4\left(Im-I1\right)}{R1}\left(A\right)\)
\(\Rightarrow Im=\dfrac{18}{Rtd}=\dfrac{18}{\dfrac{R2R3}{R2+R3}+\dfrac{R1R4}{R1+R4}}=\dfrac{18}{10+\dfrac{400}{R1+R4}}\left(A\right)\)
\(\Rightarrow I1=\dfrac{U-U23}{R1}=\dfrac{18-Im.R23}{R1}=\dfrac{18-\dfrac{180}{10+\dfrac{400}{R1+R4}}}{R1}=\dfrac{\dfrac{180+\dfrac{7200}{R1+R4}-180}{10+\dfrac{400}{R1+R4}}}{R1}=\dfrac{\dfrac{\dfrac{7200}{R1+R4}}{\dfrac{10R1+10R4+400}{R1+R4}}}{R1}=\dfrac{\dfrac{7200}{10R1+10R4+400}}{R1}=\dfrac{7200}{R1\left(10R1+10R4+400\right)}=\dfrac{7200}{10R1^2+400R1+4000}\left(A\right)\)
\(\Rightarrow I3+I2=Im\Rightarrow I3=\dfrac{Im}{2}=\dfrac{\dfrac{18}{10+\dfrac{400}{R1+R4}}}{2}=\dfrac{9}{10+\dfrac{400}{R1+R4}}\left(A\right)\)
\(\Rightarrow\dfrac{9}{10+\dfrac{400}{R1+R4}}-\dfrac{7200}{10R1^2+400R1+4000}=0,3\Rightarrow\dfrac{9}{10+\dfrac{400}{R1+\dfrac{400}{R1}}}-\dfrac{7200}{10R1^2+400R1+4000}=0,3\Rightarrow\left\{{}\begin{matrix}R1=40\Omega\\R4=10\Omega\end{matrix}\right.\)\(\)
26 tháng 8 2018
Hiệu điện thế giữa 2 đầu điện trở R1 là :
\(U_1=I_1.R_1=2,2.25=55\left(V\right)\)
Hiệu điện thế giữa 2 đầu điện trở R2 là :
\(U_2=I_2.R_2=1,5.30=45\left(V\right)\)
Nếu R1//R2 thì :
\(U=U_1=U_2\)
=> U = 15V thì 2 điện trở ko phù hợp để mắc song song.
