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Trong đó R1=2 ôm, R2=6 ôm, R3=4 ôm, R4=10 ôm. Hiệu điện thế Uab=28v
Tóm tắt:
\(R_1=2\Omega\\ R_2=6\Omega\\ R_3=4\Omega\\ R_4=10\Omega\\ U_{AB}=28V\\ a,R_{tđ}=?\\ b,I_1?\\ I_2=?\\ I_3=?\\ I_4=?\\ c,U_1=?\\ U_2=?\)
Giải:
Cấu tạo: R1nt[R4//(R2ntR3)]
\(R_{23}=R_2+R_3=6+4=10\left(\Omega\right)\)
\(R_{234}=\dfrac{R_4\cdot R_{23}}{R_4+R_{23}}=5\Omega\)
\(R_{tđ}=R_1+R_{234}=7\Omega\)
b,\(I_1=I=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{28}{7}=4\left(A\right)\)
\(I_{234}=I_1=4\left(A\right)\)
\(\Rightarrow U_{23}=U_4=U_{234}=I_{234}\cdot R_{234}=4\cdot5=20\left(V\right)\)
\(I_4=\dfrac{U_4}{R_4}=\dfrac{20}{10}=2\left(A\right)\)
\(I_2=I_3=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{20}{10}=2\left(A\right)\)
c,\(U_1=R_1\cdot I_1=2\cdot4=8\left(V\right)\)
\(U_2=I_2\cdot R_2=2\cdot6=12\left(V\right)\)
a)CTM: \(R_1nt\left(\left(R_2ntR_3\right)//R_4\right)\)
\(R_{23}=R_2+R_3=7+5=12\Omega\)
\(R_{234}=\dfrac{R_{23}\cdot R_4}{R_{23}+R_4}=\dfrac{12\cdot11}{12+11}=\dfrac{132}{23}\Omega\)
\(R_{tđ}=R_1+R_{234}=3+\dfrac{132}{23}=\dfrac{201}{23}\Omega\)
b)\(I_1=I_{234}=I_{AB}=\dfrac{U_{AB}}{R_{AB}}=\dfrac{30}{\dfrac{201}{23}}=\dfrac{230}{67}A\approx3,4A\)
\(U_{23}=U_4=U-U_1=30-I_1\cdot R_1=30-\dfrac{230}{67}\cdot3=\dfrac{1320}{67}V\)
\(I_4=\dfrac{U_4}{R_4}=\dfrac{\dfrac{1320}{67}}{11}=\dfrac{120}{67}A\approx1,79A\)
\(I_2=I_3=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{\dfrac{1320}{67}}{12}=\dfrac{110}{67}A\approx1,64A\)
MCD: R1 nt(R2//R3)
a, ĐIện trở tương đương của đoạn mạch
\(R_{23}=\dfrac{R_2R_3}{R_2+R_3}=\dfrac{30\cdot20}{30+20}=12\left(\Omega\right)\)
\(R_{tđ}=R_1+R_{23}=18+12=30\left(\Omega\right)\)
b,Cường độ dòng điện qua mỗi điện trở
\(I_1=I_{23}=I=\dfrac{U}{R_{tđ}}=\dfrac{60}{30}=2\left(A\right)\)
\(U_2=U_3=U_{23}=I_{23}\cdot R_{23}=2\cdot12=24\left(V\right)\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{24}{30}=0,8\left(A\right)\)
\(I_3=\dfrac{U_3}{R_3}=\dfrac{24}{20}=1,2\left(A\right)\)
1. a. Theo ht 4' trg đm //, ta có: Rtđ= (R1.R2)/(R1+R2)= (3.6)/(3+6)=2 ôm
b.Theo ĐL ôm, ta có: I= U/Rtđ=24/2=12 A
I1=U/R1=24/3=8 ôm
I2=U/R2=24/6=4 ôm
2. a. Theo ht 4' trg đm //, ta có: Rtđ=(R1.R2.R3)/(R1+R2+R3)= (6.12.4)/(6+12+4)=13,09 ôm
b. Áp dụng ĐL Ôm, ta có: U=I.R=3.13,09=39,27 V
c. Theo ĐL Ôm, ta có:
I1=U/R1=39,27/6=6.545 A
I2=U/R2=39,27/12=3,2725 A
I3=U/R3=39,27/4=9.8175 A
a, \(=>R1//R2//R3//R4\)
\(=>\dfrac{1}{Rtđ}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}+\dfrac{1}{R4}=\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{20}\)
\(=>Rtd=\dfrac{10}{3}\left(om\right)\)
b, \(=>U=U1=U2=U3=U4=24V\)
\(=>I1=\dfrac{U1}{R1}=\dfrac{24}{10}=2,4A\)
\(=>I2=\dfrac{U2}{R2}=\dfrac{24}{10}=2,4A\)
\(=>I3=\dfrac{U3}{R3}=\dfrac{24}{20}=1,2A\)
\(=>I4=\dfrac{U4}{R4}=\dfrac{24}{20}=1,2A\)
a) Ta có (R1ntR2)//(R3ntR4)
Đặt x là R4
=> Rtđ=\(\dfrac{R12.R34}{R12+R34}=\dfrac{18.\left(12+x\right)}{18+12+x}=\dfrac{216+18x}{30+x}\)
=> \(I=\dfrac{U}{Rt\text{đ}}=36:\dfrac{216+18x}{30+x}=\dfrac{36.\left(30+x\right)}{216+18x}=\dfrac{36.\left(30+x\right)}{18\left(12+x\right)}=\dfrac{2.\left(30+x\right)}{12+x}=\dfrac{60+2x}{12+x}\)
Vì R12//R34=>U12=U34=U=36V
Vì R1ntR2=>I1=I2=I12=\(\dfrac{U12}{R12}=\dfrac{36}{18}=2A\)(1)
Vì R3ntR4=>I3=I4=I34=\(\dfrac{U34}{R34}=\dfrac{36}{12+x}\)(2)
Mặt khác theo đề ra I1=I2=I3=I4 mà I1=I2=2A=>I1=I2=I3=I4=2A
=> \(I4=\dfrac{36}{12+x}=2=>x=6\Omega\)
=> R4= 6 \(\Omega\)
b) Ta có R1ntR2)//(R3ntR4)
=> Rtđ=\(\dfrac{R12.R34}{R12+R34}=12\Omega\)
=> \(I=\dfrac{U}{Rt\text{đ}}=\dfrac{36}{12}=3A\)
Vì R12//R34=> U12=U34=U=36V
Vì R3ntR4=> I3=I4=I34=\(\dfrac{U34}{R34}=\dfrac{36}{36}=1A\)
vì R1ntR2=>I1=I2=I12=\(\dfrac{U12}{R12}=\dfrac{36}{18}=2A\)
=> Ucd=-Uad+Uac=-U3+U1=-(I3.R3).(I1.R1)=-12+24=12V