Question

Cho hình vẽ . Có U_AB = 18V , ampe kế A₁ chỉ 2A , ampe kế A chỉ 4.5 A . Tính R₁ , R₂ , R₃ biết R₃ = 4R₂.

mọi người giúp mik nha !!!! Cảm ơn nhìu !!!!

Answer 1

Ta có R1//R2//R3=>U1=U2=U3=U=18V

=>R1=\(\dfrac{U1}{Ia1}=\dfrac{18}{2}=9\Omega\)

Ta có Ia=Ia1+I23=>I23=Ia-Ia1=2,5A

Ta có R2//R3=>U2=U3=U23=I23.R3=2,5.\(\dfrac{R2.R3}{R2+R3}=\dfrac{R2.4R2}{R2+4R2}\).2,5=18 (1)

Giải pt 1 => R2=9\(\Omega\)

R3=4R2=36\(\Omega\)

Answer 2

R1//R2//R3

=> \(R_{td}=\dfrac{U_{AB}}{I}=\dfrac{18}{4,5}=4\Omega\)

\(U_{AB}=U_1=18\left(V\right)\)

\(R_1=\dfrac{U_1}{I_1}=\dfrac{18}{2}=9\Omega\)

\(\dfrac{1}{R_{td}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\)

\(\Rightarrow\dfrac{1}{R_{td}}-\dfrac{1}{R_1}=\dfrac{1}{R_2}+\dfrac{1}{R_3}\)

\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{9}=\dfrac{1}{R_2}+\dfrac{1}{3R_2}\)

\(\Leftrightarrow\dfrac{5}{36}=\dfrac{4}{3R_2}\)

\(\Rightarrow R_2=9,6\Omega\)

\(R_3=3R_2=3.9,6=28,8\Omega\)