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ý là thế này hả bn?
(R1ntR2)//(R3ntR4)
a,\(=>Rtd=\dfrac{\left(R1+R2\right)\left(R3+R4\right)}{R1+R2+R3+R4}=\dfrac{\left(10+15\right)\left(10+25\right)}{10+15+10+25}=\dfrac{175}{12}\left(om\right)\)
b,\(=>U12=U34=36V\)
\(=>I12=I1=I2=\dfrac{U12}{R12}=\dfrac{36}{10+15}=1,44A\)
\(=>I34=I3=I4=\dfrac{U34}{R34}=\dfrac{36}{10+25}=\dfrac{36}{35}A\)
ta có sơ đồ:
Ta có: R12=\(\dfrac{R_1R_2}{R_1+R_2}=\dfrac{10.20}{10+20}=\dfrac{200}{30}=\dfrac{20}{3}\left(\Omega\right)\)
R123=R12+R3=\(\dfrac{20}{3}+30=\dfrac{110}{3}\left(\Omega\right)\)
=> Rtd=R1234=\(\dfrac{R_{123}R_4}{R_{123}+R_4}=\dfrac{\dfrac{110}{3}.40}{\dfrac{110}{3}+40}=\dfrac{440}{23}=19,13\left(\Omega\right)\)
=> I=\(\dfrac{U}{R_{td}}=\dfrac{90}{\dfrac{440}{23}}=\dfrac{207}{44}=4,7\left(A\right)\)
Lại có:
U=U4=U123=90(V)
=> I4=U4:R4=90:40=2,25(A)
I12=I3=U123:R123=\(\dfrac{90}{\dfrac{110}{3}}=2,45\left(A\right)\)
U12=U1=U2=U-U3=U-I3R3=90-\(\dfrac{27}{11}.30\)=\(\dfrac{180}{11}=16,36\left(V\right)\)
=> I1=\(\dfrac{U_1}{R_1}=\dfrac{\dfrac{180}{11}}{10}=\dfrac{18}{11}=1,636\left(A\right)\)
I2\(=\dfrac{U_2}{R_2}=\dfrac{\dfrac{180}{11}}{20}=\dfrac{9}{11}=0,818\left(A\right)\)
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)
\(\Rightarrow U_1=U_2=U_{12}=I_{12}.R_{12}=0,5.20=10\left(V\right)\)
\(\Rightarrow I_1=\frac{U_1}{R_1}=\frac{10}{30}=\frac{1}{3}\left(A\right)\)
\(\Rightarrow I_2=0,5-\frac{1}{3}=\frac{1}{6}\left(A\right)\)
\(\Rightarrow U_3=U_4=U_{34}=I_{34}.R_{34}=0,5.24=12\left(V\right)\)
\(\Rightarrow I_3=\frac{U_3}{R_3}=\frac{12}{40}=0,3\left(A\right)\)
\(\Rightarrow I_4=\frac{U_4}{R_4}=\frac{12}{60}=0,2\left(A\right)\)
Đáp án D
Giữa I 1 , I 2 , I 3 có mối liên hệ là I 2 = I 3 = I 1 / 2
a)Ta có (R1//R3)nt(R2//R4)=> Rtđ=R13+R24=\(\dfrac{R1.R3}{R1+R3}+\dfrac{R2.R4}{R2+R4}=1+2=3\Omega\)
=> I=\(\dfrac{U}{Rt\text{đ}}=\dfrac{5}{3}A\)
Vì R13ntR24=>I13=I24=I=\(\dfrac{5}{3}A\)
Vì R1//R3=> U1=U3=U13=I13.R13=\(\dfrac{5}{3}.1=\dfrac{5}{3}V\)
=> I1=\(\dfrac{U1}{R1}=\dfrac{5}{3}:2=\dfrac{5}{6}A;I3=\dfrac{U3}{R3}=\dfrac{5}{3}:2=\dfrac{5}{6}A\)
Vì R2//R4=> U2=U4=U24=I24.R24=\(\dfrac{5}{3}.2=\dfrac{10}{3}V\)
=> I2=\(\dfrac{U2}{R2}=\dfrac{10}{3}:3=\dfrac{10}{9}A;I4=\dfrac{U4}{R4}=\dfrac{10}{3}:6=\dfrac{5}{9}A\)
Vì I1<I2=> Chốt dương tại D
=> I1+Ia=I2=> Ia=I2-I1=\(\dfrac{5}{18}A\)
Vậy ampe kế chỉ 5/18 A
Giải
a. Do \(R_1\)//\(R_2\) nên :
\(R_{12}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{20.20}{20+20}=10\Omega\)
\(R_3\) nt \(\left(R_1//R_2\right)\) nên điện trở tương đương là :
\(R_{tđ}=R_{12}+R_3=10+5=15\Omega\)
b. CĐDĐ qua mạch chính là :
\(I=\dfrac{U}{R}=\dfrac{15}{15}=1A\)
Vì \(R_{12}\) nt \(R_3\) nên :
\(I=I_3=I_{12}=1A\)
\(\Rightarrow U_{12}=I_{12}.R_{12}=1.10=10V\)
Vì \(R_1//R_2\) nên :
\(U_{12}=U_1=U_2=10V\)
CĐDĐ qua mỗi ĐT là :
\(I_1=\dfrac{U_1}{R_1}=\dfrac{10}{20}=0,5A\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{10}{20}=0,5A\)
*: \(R1ntR2ntR3=>RTđ=R1+R2+R3=\dfrac{U}{I}=\dfrac{110}{2}=55\left(ôm\right)\)(1)
**: \(R1ntR2=>Rtđ=R1+R2=\dfrac{U}{I1}=\dfrac{110}{5,5}=20\left(ôm\right)\)
\(=>R2=20-R1\left(2\right)\)
*** \(R1ntR3=>Rtđ=R1+R3=\dfrac{U}{I2}=\dfrac{110}{2,2}=50\left(ôm\right)\)
\(=>R3=50-R1\left(3\right)\)
(1)(2)(3)
\(=>R1+20-R1+50-R1=55=>R1=15\left(\cdotôm\right)\)
\(=>R2=20-R1=5\left(om\right)\)
\(=>R3=50-R1=35\left(ôm\right)\)
a, \(=>R5nt\left[\left(R1//R3\right)nt\left(R2//R4\right)\right]\)
\(=>Rtd=R5+\dfrac{R1R3}{R1+R3}+\dfrac{R2.R4}{R2+R4}=1,2+\dfrac{8.12}{8+12}+\dfrac{8.24}{8+24}\)\(=12\left(om\right)\)
b,\(=>Im=\dfrac{4}{12}=\dfrac{1}{3}A=I5=I13=I24\)
\(=>U13=\dfrac{1}{3}\left(\dfrac{8.12}{8+12}\right)=1,6V=U1=U3\)
\(=>I1=\dfrac{1,6}{8}=0,2A,I3=\dfrac{1,6}{12}=\dfrac{2}{15}A\)
\(=>U24=\dfrac{1}{3}\left(\dfrac{8,24}{8+24}\right)=2V=U2=U4\)
\(=>I2=\dfrac{2}{8}=0,25A,I4=\dfrac{2}{24}=\dfrac{1}{12}A\)
c, giả sử chiều dòng điện qua ampe kế từ M đến N
\(=>I1=Ia+I2=>Ia=I1-I2=0,2-0,25=-0,05A\)
=>chiều dòng điện phải từ N tới M =>số chỉ ampe kế là 0,05A
`[(R_1 nt R_2 ) //// R_3]nt R_4`
`I=I_4=I_[123]=1,5(A)`
`U_[123]=I_[123].R_[123]=1,5. 1/[1/20 +1/[10+10]]=15(V)=U_3=U_[12]`
`=>I_3=[U_3]/[R_3]=15/20=0,75(A)`
`I_[12]=[U_[12]]/[R_[12]]=15/[10+10]=0,75(A)=I_1=I_2`