a, K mở : (R5//R3) nt R2 nt R1

I_a = I₅ = 1.125 A

⇒U₅ = I₅.R₅ = 9V

U₁ + U₂ = U - U₅

⇔IR₁ + IR₂ = 27

⇒I = 1.5 A

I₃ = I - I₅ = 0.375 A

⇒ R_{3 = \(\dfrac{U_3}{R_3}\)} = 24 Ω

Bài dễ mà bn, ADCT là ra :))

Có \(R_{tđ}=\frac{R_1.R_2}{R_1+R_2}+\frac{R_3.R_4}{R_3+R_4}=\frac{30.60}{30+60}+\frac{60.R_3}{60+R_3}\)\(=20+\frac{60.R_3}{60+R_3}\)

Có \(R_{tđ}=\frac{U_{AB}}{I}=\frac{22}{0,5}=44\left(\Omega\right)\)

\(\Rightarrow20+\frac{60R_3}{60+R_3}=44\Leftrightarrow\frac{60R_3}{60+R_3}=24\)

\(\Leftrightarrow R_3=40\left(\Omega\right)\)

b/ Có I=I₁₂=I₃₄= 0,5(A)

\(\Rightarrow U_1=U_2=U_{12}=I_{12}.R_{12}=0,5.20=10\left(V\right)\)

\(\Rightarrow I_1=\frac{U_1}{R_1}=\frac{10}{30}=\frac{1}{3}\left(A\right)\)

\(\Rightarrow I_2=0,5-\frac{1}{3}=\frac{1}{6}\left(A\right)\)

\(\Rightarrow U_3=U_4=U_{34}=I_{34}.R_{34}=0,5.24=12\left(V\right)\)

\(\Rightarrow I_3=\frac{U_3}{R_3}=\frac{12}{40}=0,3\left(A\right)\)

\(\Rightarrow I_4=\frac{U_4}{R_4}=\frac{12}{60}=0,2\left(A\right)\)

Xin lỗi mình ko thấy hình ạ!

bạn có thể vt ra MCD đc ko?

a)Ta có (R1//R3)nt(R2//R4)=> Rtđ=R13+R24=\(\dfrac{R1.R3}{R1+R3}+\dfrac{R2.R4}{R2+R4}=1+2=3\Omega\)

=> I=\(\dfrac{U}{Rt\text{đ}}=\dfrac{5}{3}A\)

Vì R13ntR24=>I13=I24=I=\(\dfrac{5}{3}A\)

Vì R1//R3=> U1=U3=U13=I13.R13=\(\dfrac{5}{3}.1=\dfrac{5}{3}V\)

=> I1=\(\dfrac{U1}{R1}=\dfrac{5}{3}:2=\dfrac{5}{6}A;I3=\dfrac{U3}{R3}=\dfrac{5}{3}:2=\dfrac{5}{6}A\)

Vì R2//R4=> U2=U4=U24=I24.R24=\(\dfrac{5}{3}.2=\dfrac{10}{3}V\)

=> I2=\(\dfrac{U2}{R2}=\dfrac{10}{3}:3=\dfrac{10}{9}A;I4=\dfrac{U4}{R4}=\dfrac{10}{3}:6=\dfrac{5}{9}A\)

Vì I1<I2=> Chốt dương tại D

=> I1+Ia=I2=> Ia=I2-I1=\(\dfrac{5}{18}A\)

Vậy ampe kế chỉ 5/18 A

mạch R1nt((R2ntR3)//R4)

=> RTđ= R1+R234=4+\(\dfrac{\left(R2+R3\right).R4}{R2+R3+R4}=6,4\Omega\)

=> I=\(\dfrac{U}{Rt\text{đ}}=\dfrac{12}{6,4}=1,875A\)

Vì R1ntR234=>I1=I234=I=1,875A

Vì R23//R4=>U23=U4=U234=I234.R234=1,875.2,4=4,5V

=>I4=\(\dfrac{U4}{R4}=\dfrac{4,5}{4}=1,125A\)

Vì R2ntR3=>I2=I3=I23=\(\dfrac{U23}{R23}=\dfrac{4,5}{6}=0,75A\)

thiếu mạch

Mạch như nào bạn ? 

R₁₂= \(\frac{R1.R2}{R1+R2}=\frac{3.6}{3+6}=2\left(\Omega\right)\)

R123=R12+R3=2+8=10(Ω)

R1234=\(\frac{R123.R4}{R123+R4}=\frac{10.10}{10+10}=5\left(\Omega\right)\)

R_tđ=R1234+R5=5+5=10(Ω)

I_mạch=\(\frac{U_{ab}}{R_{tđ}}=\frac{100}{10}=10\left(A\right)=I_5\)

I₄=\(\frac{R123}{R123+R4}.I_{mạch}=\frac{10}{10+10}.10=5\left(A\right)\)

I₁₂₃=I_mạch-I₄=10-5=5(A)=I₃

I₁=\(\frac{R2}{R1+R2}.I_{123}=\frac{6}{3+6}.5=\frac{10}{3}\left(A\right)\)

I₂=I₁₂₃-I₁=5-\(\frac{10}{3}\)= \(\frac{5}{3}\left(A\right)\)

Rn=[(R1+R2)*(R3+R4)]/[(R3+R4)]=THẾ SỐ=4(ÔM)

I_c=E/(r+Rn)=60/(2+4)=10A

Uab=Rn*Ic=4*10=40(v)

I₁=I₂=Uab/(R1+R2)=40/6=20/3A

I₃=I_{4=Uab/(R3+R4)=40/(12)=10/3A}

_{Png=EI=60*10=600((w)}

sẵn cho mình hỏi làm sao bạn gửi câu hỏi được vậy?