a) Ta có (R2ntR3)//R4)ntR1=>R23=10 \(\Omega\);R234=\(\dfrac{10.40}{10+40}=8\Omega=>Rtđ=R234+R1=28\Omega\)
=>\(I=I1=I234=\dfrac{U}{Rtđ}=2,6A\)=>U1=I1.R1=52V
Vì R23//R4=>U23=U4=U234=I234.R234=2,6.8=20,8V=>\(I4=\dfrac{U4}{R4}=0,52A\)
Vì R2ntR3=>I2=I3=I23=\(\dfrac{U23}{R23}=2,08A\)=>U2=I2.R2=12,48V;U3=I3.R3=8,32V
Vậy.....
Tóm tắt:
\(R_1=20\Omega\)
\(R_2=6\Omega\)
\(R_3=4\Omega\)
\(R_4=40\Omega\)
\(U_{AB}=72,8V\)
\(R_{TĐ}=?\)
\(I_1,I_2,I_3,I_4=?\)
\(U_1,U_2,U_3,U_4=?\)
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Bài làm:
a) - Sơ đồ mạch điện: \(R_1nt\left[\left(R_2ntR_3\right)\text{/}\text{/}R_4\right]\)
Từ sơ đồ mạch điện: \(\Rightarrow R_{23}=R_2+R_3=6+4=10\left(\Omega\right)\)
\(\Rightarrow R_{234}=\dfrac{R_{23}\cdot R_4}{R_{23}+R_4}=\dfrac{10\cdot40}{10+40}=8\left(\Omega\right)\)
\(\Rightarrow R_{TĐ}=R_1+R_{234}=20+8=28\left(\Omega\right)\)
b) Cường độ chạy qua đoạn mạch AB là:
\(I_{AB}=\dfrac{U_{AB}}{R_{TĐ}}=\dfrac{72,8}{28}=2,6\left(A\right)\)
Vì \(R_1ntR_{234}\) nên: \(I_1=I_{234}=I_{AB}=2,6\left(A\right)\)
\(\Rightarrow U_1=I_1\cdot R_1=2,6\cdot20=52\left(V\right)\)
Vì \(R_{23}\text{/}\text{/}R_4\) nên :\(U_{23}=U_4=U_{234}=I_{234}\cdot R_{234}=2,6\cdot8=20,8\left(V\right)\)
\(\Rightarrow I_4=\dfrac{U_4}{R_4}=\dfrac{20,8}{40}=0,52\left(A\right)\)
Vì \(R_2ntR_3\) nên: \(I_2=I_3=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{20,8}{10}=2,08\left(A\right)\)
\(\Rightarrow U_2=I_2\cdot R_2=2,08\cdot6=12,48\left(V\right)\)
\(\Rightarrow U_3=I_3\cdot R_3=2,08\cdot4=8,32\left(V\right)\)
Vậy .....................................
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