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: \(\dfrac{1}{R_{td}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\Leftrightarrow\dfrac{1}{12}=\dfrac{1}{30}+\dfrac{1}{R_2}\Leftrightarrow R_2=20\Omega\)

\(\Rightarrow\left[{}\begin{matrix}A_1=\dfrac{U_1}{R_1}=\dfrac{48}{30}=1,6A\\A_2=\dfrac{U_2}{R_2}=\dfrac{48}{20}=2,4A\end{matrix}\right.\)

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.\(=>U=\left(Ia1+Ia2\right).Rtd=\left(1+1,5\right).22=55V\)

\(Ia=I12=4A\)

\(Ia1=I1=1A\)

\(=>Ia2=Ia-Ia1=3A\)

\(=>Uv=U2=U1=I2.R2=60\left(V\right)=>R1=\dfrac{U1}{I1}=60\left(om\right)\)

a, ta có I1+I2=I=3(A)

\(I_1=\dfrac{36}{30}=1,2\left(A\right)\)

\(\Rightarrow I_2=3-I_1=1,8\left(A\right)\)

b, \(R_2=\dfrac{U}{I_2}=\dfrac{36}{1,8}=20\left(\Omega\right)\)

\(a,=>R1//R2\)

\(=>Ia=I1+I2=3A\)

\(=>Uv=U1=U2=36V\)

\(=>I1=\dfrac{U1}{R1}=\dfrac{36}{30}=1,2A=Ia1\)

\(=>I2=I1-I1=3-1,2=1,8A=Ia2\)

b, \(=>Rtd=\dfrac{30R2}{30+R2}=\dfrac{U}{Ia}=\dfrac{36}{3}=12=>R2=20\left(om\right)\)

a/ \(R_m=R_1+\frac{R_2.R_3}{R_2+R_3}=4+\frac{10.15}{10+15}=10\left(Ôm\right)\)

b/ \(I_m=\frac{U}{R_m}=\frac{5}{10}=\frac{1}{2}\left(A\right)=I_{23}=I_1\)

\(U_{23}=I_{23}.R_{23}=2\left(V\right)=U_2=U_3\)

\(I_2=\frac{U_2}{R_2}=\frac{2}{10}=\frac{1}{5}\left(A\right)\)

\(I_3=\frac{U_3}{R_3}=\frac{2}{15}\left(A\right)\)

Vậy....