a)CTM: \(R_1//R_2//R_3\)

\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{1}{2}\Rightarrow R_{tđ}=2\Omega\)

\(U_1=U_2=U_3=U=4,8V\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{4,8}{4}=1,2A\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{4,8}{6}=0,8A\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{4,8}{12}=0,4A\)

b)CTM: \((R_1//R_2//R_3)ntR_4\)

\(I_4=I_{123}=I_{AB}=1A\)

\(R_{tđ}=\dfrac{U_{AB}}{I_{AB}}=\dfrac{4,8}{1}=4,8\Omega\)

\(R_4=R_{tđ}-R_{123}=4,8-2=2,8\Omega\)

a) Vì R1//R2 nên: \(\frac{1}{R12}\)=\(\frac{1}{R1}\)+\(\frac{1}{R2}\)= 1/6+1/12= 1/4 => R12= 4(\(\Omega\))

Vì R3 nt R12 nên: Rtđ= R3 + R12 = 16 + 4 = 20 (\(\Omega\))

b) CĐDĐ qua mạch chính là: I= U/Rtđ= 30/20= 1,5(A)

TRong mạch song² : \(\frac{I1}{I2}\)= \(\frac{R2}{R1}\)= \(\frac{12}{6}\)=2 \(\Leftrightarrow\) I₁=2I₂

Vì R₃ nt R₁₂ nên: I = I₁₂=I₃ = 1,5(A)

Mà: R₁₂= R₁+R₂=> R₁₂= 2R₂ + R₂ = 3R₂

3R₂ = 1,5A => R₂= 0,5(A)

\(\Leftrightarrow\)R₁= 2R₂= 0,5 . 2= 1(A)

a/ R=20

b/ I=1,5A

Bài 3:

a. Cần mắc vào HĐT 220V để sáng bình thường.

b. \(I=P:U=1100:220=5A\)

c. \(A=Pt=1100.2.30=66000\)Wh = 66kWh = 237 600 000J

d. \(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{p}=\dfrac{\left(220:5\right).0,45.10^{-6}}{1,10.10^{-6}}=18\left(m\right)\)

Bài 4:

a. \(Q_{toa}=A=I^2Rt=2,4^2\cdot120\cdot25=17280\left(J\right)\)

b. \(Q_{thu}=mc\Delta t=1.4200.75=315000\left(J\right)\)

\(H=\dfrac{Q_{thu}}{Q_{toa}}100\%=\dfrac{17280}{315000}100\%\approx5,5\%\)

Baì 1:

a. \(R=R1+R2=4+6=10\Omega\)

\(I=I1=I2=U:R=18:10=1,8A\left(R1ntR2\right)\)

b. \(R1nt\left(R2\backslash\backslash\mathbb{R}3\right)\)

 \(R'=R1+\left(\dfrac{R2.R3}{R2+R3}\right)=4+\left(\dfrac{6.12}{6+12}\right)=8\Omega\)

\(I'=U:R'=18:8=2,25A\)

Bài 2:

a. \(R=\dfrac{R1.R2}{R1+R2}=\dfrac{15.10}{15+10}=6\Omega\)

b. \(U=U1=U2=18V\left(R1\backslash\backslash\mathbb{R}2\right)\)

\(\Rightarrow\left\{{}\begin{matrix}I1=U1:R1=18:15=1,2A\\I2=U2:R2=18:10=1,8A\end{matrix}\right.\)

Mình tính ra là 54,25% ý.

R1 mắc nối tiếp với r2 ạ

\(a.R_{tđ}=R_1+R_2=4+6=10\Omega\\ b.R_{tđ}'=R_1+\dfrac{R_2.R_3}{R_2+R_3}=4+\dfrac{6.12}{6+12}=8\Omega\\ I=\dfrac{U_{AB}}{R_{tđ}'}=\dfrac{18}{8}=2,25A\\ Vì.R_1ntR_{23}\\ \Rightarrow I=I_1=I_{23}=2,25A\\ U_1=I_1.R_1=4.2,25=9V\\ U_{23}=U_{AB}-U_1=18-9=9V\\ Vì.R_2//R_3\Rightarrow U_{23}=U_2=U_3=9V\\ I_3=\dfrac{U_3}{R_3}=\dfrac{9}{12}=0,75A\)

a) \(R_1ntR_2\Rightarrow R_{tđ}=R_1+R_2=4+6=10\Omega\)

\(I_m=\dfrac{U}{R_{tđ}}=\dfrac{18}{10}=1,8A\)

b) CTM: \(R_1nt\left(R_2//R_3\right)\)

\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{6\cdot12}{6+12}=4\Omega\)

\(R_{tđ}=R_1+R_{23}=4+4=8\Omega\)

c)\(I_m=\dfrac{U}{R_{tđ}}=\dfrac{18}{8}=2,25A\)

\(R_1nt\left(R_2//R_3\right)\Rightarrow I_{23}=I_1=I_m=2,25A\)

\(U_{23}=I_{23}\cdot R_{23}=2,25\cdot4=9V\Rightarrow U_3=9V\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{9}{12}=0,75A\)

vl hỏi đ ai trả lời, quê

Ôi con sông quê! Con sông quê!

a, \(=>R1//R2//R3//R4\)

\(=>\dfrac{1}{Rtđ}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}+\dfrac{1}{R4}=\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{20}\)

\(=>Rtd=\dfrac{10}{3}\left(om\right)\)

b, \(=>U=U1=U2=U3=U4=24V\)

\(=>I1=\dfrac{U1}{R1}=\dfrac{24}{10}=2,4A\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{24}{10}=2,4A\)

\(=>I3=\dfrac{U3}{R3}=\dfrac{24}{20}=1,2A\)

\(=>I4=\dfrac{U4}{R4}=\dfrac{24}{20}=1,2A\)

a) Điện trở tương đương của đoạn mạch:

\(Rtđ=\dfrac{R1.R2}{R1+R2}=\dfrac{15.10}{15+10}=6\left(\Omega\right)\)

b) Cường độ dòng điện chạy qua điện trở

\(I=\dfrac{U}{Rtđ}=\dfrac{18}{6}=3\left(A\right)\)

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

b)\(U_1=U_2=U=18V\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{18}{15}=1,2A;I_2=\dfrac{U_2}{R_2}=\dfrac{18}{10}=1,8A\)

c)\(R_2ntR_3\Rightarrow R_{23}=R_2+R_3=10+5=15\Omega\)

\(R_1//\left(R_2ntR_3\right)\)\(\Rightarrow R_{tđ}=\dfrac{R_1\cdot R_{23}}{R_1+R_{23}}=\dfrac{15\cdot15}{15+15}=7,5\Omega\)

\(I=\dfrac{U}{R_{tđ}}=\dfrac{18}{7,5}=2,4A\)