mỗi bài 4 thôi à bạn?

4 \(=>\left(R3//R4\right)nt\left(R1//R2\right)ntR5\)

\(=>\)\(I34=\dfrac{Uab}{Rtd}=\dfrac{6}{R5+\dfrac{R3.R4}{R3+R4}+\dfrac{R1.R2}{R1\left|R2\right|}}\)

\(=\dfrac{6}{0,5+\dfrac{3.1}{3+1}.2}=3A=I3+I4=Ia\)

chuyển mạch trên về mạch sao

\(=>x=\dfrac{R2.Rv}{R2+Rv+R4}=\dfrac{9.150}{9+150+7}=\dfrac{675}{83}\left(ôm\right)\)

\(=>y=\dfrac{R2.R4}{R2+R4+Rv}=\dfrac{9.7}{9+7+150}=\dfrac{63}{166}\left(ôm\right)\)

\(=>z=\dfrac{R4.Rv}{R4+R5+Rv}=\dfrac{150.7}{150+7+9}=\dfrac{525}{83}\left(ôm\right)\)

\(=>Rtd=\dfrac{\left(R1+x\right)\left(R3+z\right)}{R1+R3+x+z}+y=\dfrac{\left(2+\dfrac{675}{83}\right)\left(3+\dfrac{525}{83}\right)}{2+3+\dfrac{1200}{83}}\)

\(+\dfrac{63}{166}=5,2\left(ôm\right)\)

(kết quả xấu thế nhờ)

Cảm ơn nhiều ạ. 

không có hình vẽ mà tui thấy bài 4: đoạn mạnh nối tiếp nên chắc 5 điện trở nối tiếp chăng?

R1 nt R2 nt R3 nt R4 nt R5

\(=>Im=I1=I2=I3=I4=I5=\dfrac{U}{R1+R2+R3+R4+R5}\)

\(=\dfrac{5}{R1+\dfrac{R1}{2}+\dfrac{R1}{3}+\dfrac{R1}{4}+\dfrac{R1}{5}}=\dfrac{5}{\dfrac{120R1+60R1+40R1+30R1+24R1}{120}}\)

\(=\dfrac{5}{\dfrac{274R1}{120}}=\dfrac{600}{274R1}=\dfrac{300}{137R1}A\)

\(=>U1=I1.R1=\dfrac{300}{137}V\)

\(=>U2=I2R2=\dfrac{300}{137.2}=\dfrac{150}{137}V\)

\(=>U3=\dfrac{300}{137.3}=\dfrac{100}{137}V\)

\(=>U4=\dfrac{300}{137.4}=\dfrac{75}{137}V\)

\(=>U5=\dfrac{300}{137.5}=\dfrac{60}{137}V\)

3 cách mắc

c1 : 13 cái R1 nt nhau

c2: 2R2nt3R1

c3:1R2nt7R1

a, R1 nt R2

\(=>I1=I2=\dfrac{Uab}{R1+R2}=\dfrac{12}{10+5}=0,8A\)

\(=>U1=I1.R1=10.0,8=8V=>U2=Uab-U1=12-8=4V\)

b,\(\) (R1 // R3) nt R2

\(=>U2=2U1=2U3\)

\(=>U2+U1=12< =>U2+\dfrac{U2}{2}=12=>U2=8V\)

\(=>U1=\dfrac{U2}{2}=4V=U3\)

\(=>I2=I1+I3=>\dfrac{U2}{R2}=\dfrac{U1}{R1}+\dfrac{U3}{R3}\)

\(=>\dfrac{8}{5}=\dfrac{4}{10}+\dfrac{10}{R3}=>R3=8,3\left(ôm\right)\)

a,\(=>Pđ=UI=220.0,4=88W\)

b, công tơ tăng thêm 12 số \(=>A=12kWh\)

\(=>A=P.t=>12=\dfrac{88}{1000}.t=>t=136h\)

c, chưa đủ đề

c, chưa đọc kĩ dưới

\(=>\dfrac{1}{Rtd}=\dfrac{1}{R1}+\dfrac{1}{R2}=\dfrac{1}{\dfrac{220^2}{88}}+\dfrac{1}{\dfrac{220^2}{25}}=>Rtd=428\left(om\right)\)

\(=>Im=\dfrac{220}{428}=0,5A\)

a, cường độ dđ mạch 

\(I=\dfrac{U}{R_{td}}=\dfrac{12}{10+5}=0,8\left(A\right)\)

\(\Rightarrow U_1=I.R_1=8\left(V\right)\)

\(\Rightarrow U_2=I.R_2=5.0,8=4\left(V\right)\)

b, \(\Rightarrow U_1=\dfrac{4}{2}=2\left(V\right)\)

\(I=I_2=\dfrac{4}{5}=0,8\left(A\right)\)

\(I_1=\dfrac{2}{10}=0,2\left(A\right)\)

\(I_3=I_2-I_1=0,6\left(A\right)\)

\(\Rightarrow R_3=\dfrac{U_1}{I_3}=\dfrac{2}{0,6}=\dfrac{10}{3}\left(\Omega\right)\)

câu 1

a, \(R_{tđ}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{10.20}{10+20}=\dfrac{20}{3}\left(\Omega\right)\)

b, \(U_{AB}=R_2I_2=20.0,6=12\left(V\right)\)

\(I_1=\dfrac{U_{AB}}{R_1}=\dfrac{12}{10}=1,2\left(A\right)\)