a. Bạn tự vẽ sơ đồ mạch điện nhé!
\(R=\dfrac{R1.R2}{R1+R2}=\dfrac{10.5}{10+5}=\dfrac{10}{3}\left(\Omega\right)\)

b. \(U=U1=U2=12V\)(R1//R2)

\(\left[{}\begin{matrix}I1=U1:R1=12:10=1,2A\\I2=U2:R2=12:5=2,4A\end{matrix}\right.\)

\(\left[{}\begin{matrix}P1=U1.I1=12.1,2=14,4\left(W\right)\\P2=U2.I2=12.2,4=28,8\left(W\right)\end{matrix}\right.\)

c. \(\dfrac{1}{R}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}=\dfrac{1}{10}+\dfrac{1}{5}+\dfrac{1}{5}=\dfrac{1}{2}\Rightarrow R=2\left(\Omega\right)\)

\(U=U1=U2=U3=12V\)(R1//R2//R3)

\(\left[{}\begin{matrix}I1=U1:R1=12:10=1,2A\\I2=U2:R2=12:5=2,4A\\I3=U3:R3=12:5=2,4A\end{matrix}\right.\)

\(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{10.15}{10+15}=6\left(\Omega\right)\)

\(R_{tđ}=R_1+R_{23}=9+6=15\left(\Omega\right)\)

\(I=I_1=I_{23}=\dfrac{U}{R_{tđ}}=\dfrac{27}{15}=1,8\left(A\right)\)

\(U_{23}=U_2=U_3=I_{23}.R_{23}=1,8.6=10,8\left(V\right)\)

\(\left\{{}\begin{matrix}I_2=\dfrac{U_2}{R_2}=\dfrac{10,8}{10}=1,08\left(A\right)\\I_3=\dfrac{U_3}{R_3}=\dfrac{10,8}{15}=0,72\left(A\right)\end{matrix}\right.\)

\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{10\cdot15}{10+15}=6\Omega\)

\(R_m=R_1+R_{23}=R_1+\dfrac{R_2R_3}{R_2\cdot R_3}=9+\dfrac{10\cdot15}{10+15}=15\Omega\)

\(I_1=I_{23}=I_m=\dfrac{U}{R}=\dfrac{27}{15}=1,8A\)

\(U_2=U_3=U_{23}=I_{23}\cdot R_{23}=6\cdot1,8=10,8V\)

\(\Rightarrow\) \(I_2=\dfrac{U_2}{R_2}=\dfrac{10,8}{10}=1,08A\)

    \(I_3=\dfrac{U_3}{R_3}=\dfrac{10,8}{15}=0,72A\)

\(R_{tđ}=\dfrac{\left(R_1+R_2\right)R_3}{R_1+R_2+R_3}=\dfrac{\left(R+R\right)R}{R+R+R}=\dfrac{2R^2}{3R}=\dfrac{2}{3}R\)

vl hỏi đ ai trả lời, quê

Ôi con sông quê! Con sông quê!

a)

b)

* Tính \(R_{tđ}\)

\(\dfrac{1}{R_{23}}=\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{50}+\dfrac{1}{50}=\dfrac{2}{50}\left(\Omega\right)\)

=> \(R_{23}=\dfrac{50}{2}=25\left(\Omega\right)\)

\(R_{tđ}=R_{23}+R_1=25+25=50\left(\Omega\right)\)

* Tính I

\(I_{AB}=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{60}{50}=1.2\left(A\right)\)

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