Bài 4. Đoạn mạch nối tiếp

` R_(td) = R1+R2 = 48 \Omega ` 

` U_1=I_1R_1 = 0,3.12=3,6V ` 

` U_2=I_2R_2=0,3.36=10,8V ` 

` R_3=U/(I)=(24)/(0.3)=80 \Omega ` 

\(R_1\) nối tiếp \(R_2\); \(U=12V\)

a).

\(R_{tđ}=R_1+R_2=10+20=30\Omega\)

b).

\(I=\dfrac{U}{R_{tđ}}=\dfrac{12}{30}=0,4A\)

c).

\(I=I_1=I_2\)

\(\Rightarrow\left\{{}\begin{matrix}U_1=I.R_1=0,4.10=4V\\U_2=I.R_2=0,4.20=8V\end{matrix}\right.\)

a)\(R_1ntR_2\Rightarrow I_A=I_1=I_2\)

\(R_{tđ}=R_1+R_2=5+15=20\Omega\)

\(I_A=I_2=\dfrac{U_2}{R_2}=\dfrac{U_V}{R_2}=\dfrac{3}{15}=0,2A\)

b)\(U_{AB}=I\cdot R_{AB}=0,2\cdot20=4V\)

a)\(R=\rho\cdot\dfrac{l}{S}=5,5\cdot10^{-8}\cdot\dfrac{11}{0,55\cdot10^{-6}}=1,1\Omega\)

b)\(R_1=\rho\cdot\dfrac{l_1}{S_1}=5,5\cdot10^{-8}\cdot\dfrac{3}{0,55\cdot10^{-6}}=0,3\Omega\)

\(R_2=\rho\cdot\dfrac{l_2}{S}=5,5\cdot10^{-8}\cdot\dfrac{11-3}{0,55\cdot10^{-6}}=0,8\Omega\)

c)\(R'=\rho\cdot\dfrac{l'}{S}\Rightarrow40=5,5\cdot10^{-8}\cdot\dfrac{l'}{0,55\cdot10^{-6}}\)

\(\Rightarrow l'=400m\)

\(MCD:R_1ntR_2\)

Ta có: \(\dfrac{U_1}{U_2}=\dfrac{R_1}{R_2}=\dfrac{3R_2}{R_2}=3\)

\(\Rightarrow U_1=3U_2\)

Do: \(U=U_1+U_2=3U_2+U_2=4U_2\)

\(\Rightarrow U_2=\dfrac{40}{4}=10V\)

\(\Rightarrow U_1=3U_2=3\cdot10=30V\)

Bài 4.

a)Khi K đóng: \(Đ_1ntĐ_2\Rightarrow R_{tđ}=R_{Đ1}+R_{Đ2}=36+46=82\Omega\)

\(I_{Đ1}=I_{Đ2}=\dfrac{U}{R_{tđ}}=\dfrac{41}{82}=0,5A\)

b)Nếu chỉ dùng đèn 1:

\(I_{Đ1}=\dfrac{U}{R_{Đ1}}=\dfrac{41}{36}=1,134A\)