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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
R1 R2 R3 R4
a/ \(\frac{1}{R_{234}}=\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{10}+\frac{1}{6}+\frac{1}{9}=\frac{17}{45}\)
\(\Leftrightarrow R_{234}=\frac{45}{17}\left(Ôm\right)\)
\(R_m=R_1+R_{234}=5+\frac{45}{17}=\frac{130}{17}\left(Ôm\right)\)
b/ \(I_m=\frac{U}{R_m}=\frac{15}{\frac{130}{17}}=\frac{51}{26}\left(A\right)=I_1=I_{234}\)
\(U_{234}=I_{234}.R_{234}=\frac{51}{26}.\frac{45}{17}=\frac{135}{26}\left(V\right)=U_2=U_3=U_4\)
\(I_2=\frac{U_2}{R_2}=\frac{\frac{135}{26}}{10}=\frac{27}{52}\left(A\right)\)
\(I_3=\frac{U_3}{R_3}=\frac{\frac{135}{26}}{6}=\frac{45}{52}\left(A\right)\)
\(I_4=\frac{U_4}{R_4}=\frac{\frac{135}{26}}{9}=\frac{15}{26}\left(A\right)\)
Vậy...
1)
a) Vì R1 nt R2 nên :
\(U_{AB}=U_1+U_2\)
\(=>U_2=U_{AB}-U_1=12V\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{6}{30}=0,2A\)
Vì đây là mạch mắc nt nên : \(I_1=I_2=0,2A\)
\(I=\dfrac{U}{R}=>R_2=\dfrac{U_2}{I_2}=60\Omega\)
b) Vì R3//R2 => \(U_1=U_3=9V\)
\(I_{Tm}=I_1+I_2=0,4A\)
\(=>R_3=\dfrac{U_3}{I_{Tm}}=22,5\Omega\)
c) Nếu R3//R1 thì :
\(U_{tm}=U_1=9V\)
\(=>I_1=\dfrac{U_{tm}}{R_1}=0,3A\)
\(I_2=\dfrac{U_{tm}}{R_2}=0,15A\)
\(I_3=\dfrac{U_{tm}}{R_3}=0,4A\)
2)
a) Theo mạch điện ta thấy :
R1 nt R2 nên :
\(R_{tđ}=R_1+R_2=18\Omega\)
\(R_{tm}=R_{tđ}=18\Omega\)
\(I=\dfrac{U_{AB}}{R_{tm}}=0,5A\)
Vì đây là đoạn mạch nối tiếp nên :
\(I=I_1=I_2=0,5A\)
\(=>U_1=R_1.I_1=6V\)
\(=>U_2=R_2.I_2=3V\)
b) Vì R3 // R1 nên : \(U_{tm}=U_1=6V\)
\(=>R_3=\dfrac{U_{tm}}{I}=\dfrac{20}{3}\Omega\)
c) Nếu R3//R2 thì : \(U_{tm}=U_2=3V\)
\(=>R_{tđ}=\dfrac{R_2.R_3}{R_2+R_3}\approx3,16\Omega\)
\(=>I=\dfrac{U_{tm}}{R_{tm}}\approx0,95A\)
Giải:
Ta có: \(R_3//R_6\)
Nên: \(R_{36}=\dfrac{R_3.R_6}{R_3+R_6}=\dfrac{3.6}{3+6}=2\left(\Omega\right)\)
Mà: \(R_2ntR_{36}\)
Nên: \(R_{236}=R_2+R_{36}=2+2=4\left(\Omega\right)\)
\(R_5//R_{236}\) nên: \(R_{5236}=\dfrac{R_5.R_{236}}{R_5+R_{236}}=\dfrac{5.4}{5+4}=\dfrac{20}{9}\left(\Omega\right)\)
Mà: \(R_1ntR_{5236}\) nên: \(R_{15236}=R_1+R_{5236}=1+\dfrac{20}{9}=\dfrac{29}{9}\left(\Omega\right)\)
Mặt khác: \(R_4//R_{15236}\)
Nên: \(R_{td}=R_{415236}=\dfrac{R_4.R_{15236}}{R_4+R_{15236}}=\dfrac{116}{65}\left(\Omega\right)\)
Vậy: ...