Bài 5. Đoạn mạch song song

a. Ta có: \(I_x=\dfrac{U_{AB}}{R_x+R_3}=\dfrac{48}{18+12}=1,6\left(A\right)\)

\(\Rightarrow U_3=I_xR_3=1,6.12=19,2\left(V\right)\)

b. Ta có: \(I_1=2I_x\Leftrightarrow\dfrac{U_{AB}}{R_1+R_2}=2\dfrac{U_{AB}}{R_x+R_3}\Leftrightarrow\dfrac{1}{5+15}=\dfrac{2}{R_x+12}\)

\(\Rightarrow R_x=28\left(\Omega\right)\)

Ta có: \(R1ntR2\)

\(\Rightarrow I1=I2\)

Theo đề bài: \(R2=3R1\)

\(\Rightarrow I1\cdot R1=I2\cdot3R1\)

\(\Leftrightarrow U1=3U2\)

\(\Rightarrow U2=2V\)

Vậy hiệu điện thế đoạn mạch là: \(U=U1+U2=6+2=8V\)

\(MCD:R1nt\left(R2//R3//R4\right)\)

\(\Rightarrow R=R1+R_{234}=12+\left(\dfrac{1}{24}+\dfrac{1}{24}+\dfrac{1}{24}\right)=12+\dfrac{3}{24}=12+8=20\Omega\)

\(\Rightarrow I=I1=I_{234}=\dfrac{U}{R}=\dfrac{30}{20}=\dfrac{3}{2}A\)

\(\Rightarrow U_{234}=U2=U3=U4=I_{234}\cdot R_{234}=\dfrac{3}{2}\cdot8=12V\)

\(\Rightarrow I2=I3=I4=\dfrac{U2}{R2}=\dfrac{12}{24}=0,5A\)

c

c

a

c

c

IN

\(R_1//R_2\Rightarrow R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{15\cdot10}{15+10}=6\Omega\)

\(I_A=\dfrac{U}{R_{tđ}}=\dfrac{12}{6}=2A\)

\(R_1//R_2\Rightarrow U_1=U_2=U=12V\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{12}{15}=0,8A\)

\(\Rightarrow I_2=I-I_1=2-0,8=1,2A\)

Ai Giúp với

A

A nha..