Question

[Lê Đức Lương]

LÀM HỘ MIK CÂU B

Answer 1

[Nguyễn Ngọc Anh Minh]

a/

\(R_{3,4,5}=R_5+\frac{R_3R_4}{R_3+R_4}=6+\frac{10.15}{10+15}=12\Omega.\)

\(R_{2,3,4,5}=\frac{R_2R_{3,4,5}}{R_2+R_{3,4,5}}=\frac{6.12}{6+12}=4\Omega\)

\(R_{tm}=R_1+R_{2,3,4,5}=4+4=8\Omega\)

\(I_{tm}=\frac{U}{R_{tm}}=\frac{18}{8}=2,25A\)

\(U_{R_1}=I_{tm}.R_1=2,25.4=9V\)

\(U_{AB}=U-U_{R_1}=18-9=9V\)

\(I_{R_5}=\frac{U_{AB}}{R_{3,4,5}}=\frac{9}{12}=0,75A\)

\(U_{R_5}=I_5.R_5=0,75.6=4,5V\)

\(U_{R_3}=U_{R_4}=U_{AB}-U_{R_5}=9-4,5=4,5V\)

\(I_{R_3}=\frac{U_{R_3}}{R_3}=\frac{4,5}{10}=0,45A\) Vậy số chỉ trên Ampe kế A là 0,45A

\(I_{R_4}=I_{R_5}-I_{R_3}=0,75-0,45=0,35A\)

\(P_{R_4}=U_{R_4}.I_{R_4}=4,5.0,35=1,575W\)

b/

\(U_{R_3}=U_{R_4}=I_{R_3}.R_3=0,6.10=6V\)

\(I_{R_4}=\frac{U_{R_4}}{R_4}=\frac{6}{15}=0,4A\)

\(I_{R_5}=I_{R_3}+I_{R_4}=0,6+0,4=1A\)

\(U_{AB}=I_{R_5}.R_{3,4,5}=1.12=12V\)

\(U=I_{R_1}.R_1+U_{AB}\Rightarrow I_{R_1}=\frac{U-U_{AB}}{R_1}=\frac{18-12}{4}=1,5A\)

\(I_{R_2}=I_{R_1}-I_{R_5}=1,5-1=0,5A\)

\(R_2=\frac{U_{AB}}{I_{R_2}}=\frac{12}{0,5}=24\Omega\)