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 \(R_3ntR_{12}\) ⇒ \(I_3=I_{12}=2,5\left(A\right)\)

\(\Rightarrow U_3=I_3\times R_3=2,5\times10=25\left(V\right)\)

\(\Rightarrow U=U_3+U_{12}=25+30=55\left(V\right)\)

\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\\ =\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{10}\\ \Rightarrow R_{tđ}=5\Omega\)

\(U=R_{tđ}.I=5.0,6=3V\)

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)

\(R_{td}=R+\dfrac{R_1R_2}{R_1+R_2}=10+\dfrac{30.20}{30+20}=22\Omega\)

\(I=I_{12}=1,5+1=2,5\left(A\right)\)

\(U=R_{td}.I=22.2,5=55\left(V\right)\)

a) \(R_1ntR_2\)

\(U_{AB}=15V;R_1=15\Omega;R_2=10\Omega\)

\(I=?\)

\(U_1=?;U_2=?\)

BL :

\(R_{td}=R_1+R_2=25\Omega\)

\(I=\dfrac{U_{AB}}{R_{td}}=0,6\left(A\right)\)

b) \(R_3//R_1\)

\(I=1A;R_3=?\)

BL :

\(U_{AB}=U_3=15V\)

\(I_3=1-0,6=0,4A\)

\(=>R_3=\dfrac{15}{0,4}=37,5\Omega\)

c) \(R_3//R_2\)

\(I=?\)

\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{\dfrac{19}{150}}=\dfrac{150}{19}\left(\Omega\right)\)

\(I=\dfrac{U_{AB}}{R_{TM}}=\dfrac{15}{\dfrac{150}{19}}=1,9\left(A\right)\)

Vậy....................

\(MCD:R_1//R_2\Rightarrow U=U_1=U_2=48V\)

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

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

b, Có : I_A=I₁=I₂=I₃=0,4A (vì A nt R₁nt R₂nt R₃)

c, Vì V // R₂ nên : U_V=U₂=0,8V\(\Rightarrow R_2=\frac{U_2}{I_2}=\frac{0,8}{0,4}=2\left(\Omega\right)\)

Điện trở tương đương của đoạn mạch NM là :

R_tđ=R₁+R₂+R₃=R₂+(R₁+R₃)=R₂+R₂=2+2=4\(\Omega\)

Ta có : R₃=3R₁

\(\Rightarrow\)R₁+R₃=R₁+3R₁=4R₁

\(\Rightarrow2=4R_1\)

\(\Rightarrow R_1=0,5\Omega\)

\(\Rightarrow R_3=0,5.3=1,5\Omega\)

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