\(=>Im=I1=I2=I3=I4=\dfrac{U}{RTd}=\dfrac{100}{R1+\dfrac{R1}{2}+\dfrac{R1}{3}+\dfrac{R1}{4}}\)

\(=\dfrac{100}{\dfrac{24R1+12R1+8R1+6R1}{24}}=\dfrac{2400}{50R1}\left(A\right)\)

\(=>U1=I1.R1=\dfrac{2400}{50}=48V\)

\(=>U2=I2.R2=\dfrac{2400}{50R1}.\dfrac{R1}{2}=24V\)

\(=>U3=I3.R3=\dfrac{2400}{50.3}=16V\)

\(=>U4=I4.R4=\dfrac{2400}{50.4}=12V\)

theo bài ra \(=>R1ntR2ntR3ntR4\)

\(=>Rtd=R1+R2+R3+R4\)

\(=R1+\dfrac{R1}{2}+\dfrac{R1}{3}+\dfrac{R1}{4}=\dfrac{24R1+12R1+8R1+6R1}{24}\)

\(=\dfrac{50R1}{24}\left(om\right)\)

\(=>Im=\dfrac{50}{Rtd}=\dfrac{50}{\dfrac{50R1}{24}}=\dfrac{24}{R1}A=I1=I2=I3=I4\)

\(=>U1=I1.R1=\dfrac{24}{R1}.R1=24V\)

\(=>U2=I2.R2=\dfrac{24}{R1}.\dfrac{R1}{2}=12V\)

\(=>U3=I3.R3=\dfrac{24}{R1}.\dfrac{R1}{3}=8V\)

\(=>U4=I4.R4=\dfrac{24}{R1}.\dfrac{R1}{4}=6V\)

\(R_{tđ}=R_1+R_2+R_3+R_4=10R_1\left(\Omega\right)\)

\(\Rightarrow I_1=I_2=I_3=I_4=\dfrac{U}{R_{tđ}}=\dfrac{50}{10R_1}=\dfrac{5}{R_1}\)

\(\Rightarrow U_1=I_1.R_1=\dfrac{5}{R_1}.R_1=5\left(V\right)\)

\(\Rightarrow U_2=I_2.R_2=\dfrac{5}{R_1}.R_2=\dfrac{5}{R_1}.2R_1=10\left(V\right)\)

\(\Rightarrow U_3=I_3.R_3=\dfrac{5}{R_1}.R_3=\dfrac{5}{R_1}.3R_3=15\left(V\right)\)

\(\Rightarrow U_2=50-5-10-15=20\left(V\right)\)

ta có I3=\(\frac{7.5}{5}=1.5\)(A) vì mắc nối tiếp nên I1=I2=I3=1.5(A) từ đó suy ra U1,U2

\(I=I1=I2=I3=U3:R3=7,5:5=1,5A\left(R1ntR2ntR3\right)\)

\(\Rightarrow\left\{{}\begin{matrix}U1=I1\cdot R1=1,5\cdot4=6V\\U2=I2\cdot R2=1,5\cdot3=4,5V\\U=U1+U2+U3=6+4,5+7,5=18V\end{matrix}\right.\)

\(R_{tđ}=R_1+R_2+R_3=4+3+5=12\Omega\)

\(U_3=7,5V\Rightarrow I_3=1,5A\)

\(\Rightarrow I_m=I_1=I_2=I_3=1,5A\)

\(\Rightarrow U_1=1,5\cdot4=6V\)

     \(U_2=1,5\cdot3=4,5V\)

     \(U_m=U_1+U_2+U_3=6+4,5+7,5=18V\)

Phân tích đoạn mạch:  R 1   n t   ( R 2   / /   R 4 )   n t   ( R 3   / /   R 5 )

R 24 = R 2 R 4 R 2 + R 4 = 4 , 2 Ω ;   R 35 = R 3 R 5 R 3 + R 5 = 2 , 4 Ω

R = R 1   +   R 24   +   R 35 = 9 Ω ;   U 3 = U 5 = U 35 = I 3 R 3 = 8 V ; I 35 = I 24 = I 1 = I = U 35 R 35 = 10 3 A . U 24 = U 2 = U 4 = I 24 R 24 = 14 V ;   U 1 = I 1 R 1 = 8 V .