Question

Cho đoạn mạch (R1//R2) nt (R3//R4) . Biết U_AB =22V , I = 0,5A , R1=30Ω , R2=R4=60Ω . Tính

a) R3= ?

b) I1,I2,I3, I4

Answer 1

Bài dễ mà bn, ADCT là ra :))

Có \(R_{tđ}=\frac{R_1.R_2}{R_1+R_2}+\frac{R_3.R_4}{R_3+R_4}=\frac{30.60}{30+60}+\frac{60.R_3}{60+R_3}\)\(=20+\frac{60.R_3}{60+R_3}\)

Có \(R_{tđ}=\frac{U_{AB}}{I}=\frac{22}{0,5}=44\left(\Omega\right)\)

\(\Rightarrow20+\frac{60R_3}{60+R_3}=44\Leftrightarrow\frac{60R_3}{60+R_3}=24\)

\(\Leftrightarrow R_3=40\left(\Omega\right)\)

b/ Có I=I₁₂=I₃₄= 0,5(A)

\(\Rightarrow U_1=U_2=U_{12}=I_{12}.R_{12}=0,5.20=10\left(V\right)\)

\(\Rightarrow I_1=\frac{U_1}{R_1}=\frac{10}{30}=\frac{1}{3}\left(A\right)\)

\(\Rightarrow I_2=0,5-\frac{1}{3}=\frac{1}{6}\left(A\right)\)

\(\Rightarrow U_3=U_4=U_{34}=I_{34}.R_{34}=0,5.24=12\left(V\right)\)

\(\Rightarrow I_3=\frac{U_3}{R_3}=\frac{12}{40}=0,3\left(A\right)\)

\(\Rightarrow I_4=\frac{U_4}{R_4}=\frac{12}{60}=0,2\left(A\right)\)