R1ntR2ntR3 =>I=I1=I2=I3=1(A)

\(U_3=U-U_1-U_2=18-5.1-10.1=3\left(V\right)\)

\(R_3=\dfrac{U_3}{I_3}=\dfrac{3}{1}=3\Omega\)

Bai 2:

R1ntR2

\(\Rightarrow U_2=U-U_1=12-3=9\left(V\right)\)

\(I=I_1=I_2=\dfrac{U_2}{R_2}=\dfrac{9}{9}=1\left(A\right)\)

\(R_1=\dfrac{U_1}{I_1}=\dfrac{3}{1}=3\Omega\)

bài 1:

a) mạch điện nối tiếp

\(\Rightarrow R_{td}=R_1+R_2+R_3+R_4=6+6+8+4=24\Omega\)

b) R2ntR3ntR4

\(\Rightarrow R_{234}=R_{td}-R_1=24-6=18\Omega\)

\(I=I_1=I_{234}=\dfrac{U_{234}}{R_{234}}=\dfrac{12}{18}=\dfrac{2}{3}\left(A\right)\)

\(U_1=R_1.I_1=6.\dfrac{2}{3}=4\left(V\right)\)

c) \(U=U_1+U_{234}=4+12=16\left(V\right)\)

Lớp có số học sinh là:

           12.2=24(học sinh)

=>Số học sinh nữ là:

           24:(1+5).5=20(học sinh)

a) R1//R2//R3

=>\(\dfrac{1}{R_{123}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{30}+\dfrac{1}{30}+\dfrac{1}{\dfrac{30}{4}}=\dfrac{1}{5}\)

\(\Rightarrow R_{123}=5\Omega\)

R4ntR123

=> \(R_{td}=R_4+R_{123}=25+5=30\Omega\)

b) R4ntR123

=> \(I_4=I=I_{123}=\dfrac{45}{30}=1,5\left(A\right)\)

\(U_1=U_2=U_3=U_{AB}-U_4=45-1,5.25=7,5\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{7,5}{30}=0,25\left(A\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{7,5}{30}=0,25\left(A\right)\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{7,5}{\dfrac{30}{4}}=1\left(A\right)\)

600m=0,6(km) 200m=0,2(km)

Thời gian xe đạp đi đến chỗ gặp nhau là

\(t=\dfrac{0,6}{15}=0,04\left(h\right)\)

Ô tô phải đi với vận tốc là:

\(v=\dfrac{0,6+0,2}{0,04}=20\left(\dfrac{km}{h}\right)\)

R1ntR23 => I1=I23=4(A)

R2//R3

\(\Rightarrow\dfrac{I_2}{I_3}=\dfrac{R_3}{R_2}\)

\(\Leftrightarrow\dfrac{I_2}{R_3}=\dfrac{I_3}{3R_3}=\dfrac{4}{4R_3}=\dfrac{1}{R_3}\)

\(\Rightarrow\left\{{}\begin{matrix}I_2=1\left(A\right)\\I_3=3\left(A\right)\end{matrix}\right.\)

R2//R3 => U3=U2=15(V)

U1=R1.I1=10.4=40(V)

U_AB=U1+U23=40+15=55(V)

R1ntR2

\(R_{td1}=R_1+R_2=\dfrac{6}{0,4}=15\left(\Omega\right)\left(1\right)\)

R1//R2

\(R_{td2}=\dfrac{R_1R_2}{R1+R2}=\dfrac{R_1R_2}{15}=\dfrac{6}{1,8}=\dfrac{10}{3}\left(\Omega\right)\)

\(\Rightarrow R1=\dfrac{50}{R2}\left(2\right)\)

Giai (1)(2)

\(\Rightarrow R1=5\Omega\)

\(R2=10\Omega\)

Tên: Nguyễn Thị Yến Nhi

Lớp: 8-9

Link: https://hoc24.vn/vip/177259202765063

Goi t1,t2 lần lượt là thời gian đi và thời gian về

theo bài ra ta có

\(t_1.30=t_2.20\)

\(\Rightarrow\dfrac{t1}{20}=\dfrac{t2}{30}=\dfrac{1,5}{50}=0,03\)

\(\Rightarrow\left\{{}\begin{matrix}t1=0,6\left(h\right)\\t2=0,9\left(h\right)\end{matrix}\right.\)