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R nt (R1 // R2)
a,\(=>U1=U2=Ia1.R1=20.1,5=30V\)
\(=>30=Ia2.R2=>R2=30\left(ôm\right)\)
\(=>Rtd=R+\dfrac{R1R2}{R1+R2}=22\left(ôm\right)\)
b.\(=>U=\left(Ia1+Ia2\right).Rtd=\left(1+1,5\right).22=55V\)
Tóm tắt:
R3nt(R1//R2)
\(R_1=20\Omega\)
\(R_3=10\Omega\)
\(I_{A1}=I_1=1,5A\)
\(I_{A2}=I_2=1A\)
a) \(R_2=?\)
b) \(U=?\)
Bài giải:
a) \(U_1=I_1\times R_1=1,5\times20=30\left(V\right)\)
Vì R1//R2 ⇒ \(U_1=U_2=U_{12}=30\left(V\right)\)
\(\Rightarrow R_2=\frac{U_2}{I_2}=\frac{30}{1}=30\left(\Omega\right)\)
b) Ta có: \(I_{12}=I_1+I_2=1,5+1=2,5\left(A\right)\)
Vì \(R_3ntR_{12}\) ⇒ \(I_3=I_{12}=2,5\left(A\right)\)
\(\Rightarrow U_3=I_3\times R_3=2,5\times10=25\left(V\right)\)
\(\Rightarrow U=U_3+U_{12}=25+30=55\left(V\right)\)
\(MCD:R_1//R_2\Rightarrow U=U_1=U_2=48V\)
Ta có: \(A_{12}=A=4A\left(R_{12}ntA\right)\)
\(\Rightarrow R_{td}=\dfrac{U}{A_{12}}=\dfrac{48}{4}=12\Omega\)
Ta có: \(\dfrac{1}{R_{td}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\Leftrightarrow\dfrac{1}{12}=\dfrac{1}{30}+\dfrac{1}{R_2}\Leftrightarrow R_2=20\Omega\)
\(\Rightarrow\left[{}\begin{matrix}A_1=\dfrac{U_1}{R_1}=\dfrac{48}{30}=1,6A\\A_2=\dfrac{U_2}{R_2}=\dfrac{48}{20}=2,4A\end{matrix}\right.\)
a)Khóa \(K_1\) đóng, khóa \(K_2\) mở ta có CTM: \(\left(R_1ntR_2\right)//R_3\)
\(I_A=I_m=1A\)
\(R_{12}=R_1+R_2=5+5=10\Omega\)
\(R_{tđ}=\dfrac{R_{12}\cdot R_3}{R_{12}+R_3}=\dfrac{10\cdot15}{10+15}=6\Omega\)
\(U=R_{tđ}\cdot I=6\cdot1=6V=U_{12}=U_3\)
\(I_1=I_2=I_{12}=\dfrac{U_{12}}{R_{12}}=\dfrac{6}{10}=0,6A\)
\(I_3=1-0,6=0,4A\)
b)Khóa \(K_1\) mở và khóa \(K_2\) đóng ta có CTM: \(R_2//\left(R_1ntR_3\right)\)
\(R_{13}=R_1+R_3=5+15=20\Omega\)
\(R_{tđ}=\dfrac{R_2\cdot R_{13}}{R_2+R_{13}}=\dfrac{5\cdot20}{5+20}=4\Omega\)
\(I_A=\dfrac{U}{R_{tđ}}=\dfrac{6}{4}=1,5A\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{U}{R_2}=\dfrac{6}{15}=0,4A\)
\(I_1=I_3=I_{13}=I-I_2=1,5-0,4=1,1A\)
\(a,R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{5.9}{5+9}=\dfrac{45}{14}\left(\Omega\right)\)
\(R_{tđ}=R_1+R_{23}=2,4+\dfrac{45}{14}=\dfrac{393}{70}\left(\Omega\right)\)
\(b,I_m=\dfrac{U}{R_{tđ}}=\dfrac{9}{\dfrac{393}{70}}\approx1,6\left(A\right)\)
\(I_{23}=I_1=I_m=1,6\left(A\right)\)
\(U_1=I_1.R_1=1,6.2,4=3,84\left(V\right)\)
\(\rightarrow I_2=\dfrac{U}{R_2}=\dfrac{9-3,84}{5}=1,032\left(A\right)\)
Cấu tạo mạch: \(\left[\left(R_3//R_4\right)ntR_2\right]//R_1\)
\(U_1=U_{234}=U_m=24V\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{24}{12}=2A\)
\(R_{34}=\dfrac{R_3+R_4}{R_3\cdot R_4}=\dfrac{6+6}{6\cdot6}=\dfrac{1}{3}\Omega\)
\(R_{234}=R_2+R_{34}=9+\dfrac{1}{3}=\dfrac{28}{3}\Omega\)
\(I_2=I_{234}=\dfrac{U_{234}}{R_{234}}=\dfrac{24}{\dfrac{28}{3}}=\dfrac{18}{7}A\)
\(U_2=I_2\cdot R_2=\dfrac{18}{7}\cdot9=\dfrac{162}{7}V\)
\(U_{34}=I_{34}\cdot R_{34}=\dfrac{18}{7}\cdot\dfrac{1}{3}=\dfrac{6}{7}V\)
\(\Rightarrow U_3=U_{34}=\dfrac{6}{7}V\Rightarrow I_3=\dfrac{U_3}{R_3}=\dfrac{1}{7}A\)
\(I_A=I_1+I_3=2+\dfrac{1}{7}=\dfrac{15}{7}A\)
a)R1//R2
\(\dfrac{R_1}{R_2}=\dfrac{I_2}{I_1}=\dfrac{1}{1,5}=\dfrac{2}{3}\)
\(\Rightarrow3R_1=2R_2\)
\(\Leftrightarrow3.20=2R_2\)
\(\Rightarrow R_2=30\Omega\)
Rnt(R1//R2)
\(R_{td}=R+\dfrac{R_1R_2}{R_1+R_2}=10+\dfrac{30.20}{30+20}=22\Omega\)
\(I=I_{12}=1,5+1=2,5\left(A\right)\)
\(U=R_{td}.I=22.2,5=55\left(V\right)\)