Question

Answer 1

a) \(R_{234}=\left(\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}\right)^{-1}=8\left(\Omega\right)\)

\(R_{tđ}=4+8=12\left(\Omega\right)\)

\(\Rightarrow U_2=I_2\cdot R_2=0,2\cdot32=6,4\left(V\right)\)

b) \(I_1=I_{234}=\dfrac{U_2}{R_{234}}=\dfrac{6,4}{8}=0,8A\)

 

Answer 2

c, ta có từ B \(I_1=I=0,8\left(A\right)\)

\(\Rightarrow U_1=0,8.4=3,2\left(V\right)\)

\(\Rightarrow U=U_1+U_2=3,2+6,4=9,6\left(V\right)\)

 

Answer 3

\(a,\) \(=>U2=I2.R2=0,2.32=6,4V\)

b,\(R1nt\left(R2//R3//R4\right)=>\dfrac{1}{R234}=\dfrac{1}{R2}+\dfrac{1}{R3}+\dfrac{1}{R4}=>R234=8\left(om\right)\)

\(=>\)\(I1=I234=\dfrac{U2}{R234}=\dfrac{6,4}{8}=0,8A\)

c,\(=>Rtd=R1+R234=8+4=12\left(om\right)=>Uab=Rtd.Im=9,6V\)