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/\(U_{R_3}=U_{R_4}=I_{R_3}.R_3=0,6.10=6V\)\(I_{R_4}=\frac{U_{R_4}}{R_4}=\frac{6}{15}=0,4A\)\(I_{R_5}=I_{R_3}+I_{R_4}=0,6+0,4=1A\)\(U_{AB}=I_{R_5}.R_{3,4,5}=1.12=12V\)\(U=I_{R_1}.R_1+U_{AB}\Rightarrow I_{R_1}=\frac{U-U_{AB}}{R_1}=\frac{18-12}{4}=1,5A\)\(I_{R_2}=I_{R_1}-I_{R_5}=1,5-1=0,5A\)\(R_2=\frac{U_{AB}}{I_{R_2}}=\frac{12}{0,5}=24\Omega\)Câu trả lời: 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/\(U_{R_3}=U_{R_4}=I_{R_3}.R_3=0,6.10=6V\)\(I_{R_4}=\frac{U_{R_4}}{R_4}=\frac{6}{15}=0,4A\)\(I_{R_5}=I_{R_3}+I_{R_4}=0,6+0,4=1A\)\(U_{AB}=I_{R_5}.R_{3,4,5}=1.12=12V\)\(U=I_{R_1}.R_1+U_{AB}\Rightarrow I_{R_1}=\frac{U-U_{AB}}{R_1}=\frac{18-12}{4}=1,5A\)\(I_{R_2}=I_{R_1}-I_{R_5}=1,5-1=0,5A\)\(R_2=\frac{U_{AB}}{I_{R_2}}=\frac{12}{0,5}=24\Omega\)