Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :R12=R1+R2=10+10=20\(\Omega\)
Có :(R1nt R2)//R3 :
\(\Rightarrow\)R123=\(\frac{R_{12}.R_3}{R_{12}+R_3}=\frac{20.5}{20+5}=4\Omega\)
Có : R4nt(R1ntR2)//R3):
\(\Rightarrow\)Rtđ=R4+R123=6+4=10\(\Omega\)
\(\Rightarrow\)Ic=\(\frac{U}{R_{tđ}}=\frac{12}{10}=1,2A\)
\(\Rightarrow\)Ic=I4=I123=1,2A
\(\Rightarrow\)U4=I4.R4=1,2.6=7,2V
Có :R4nt((R1ntR2)//R3)
\(\Rightarrow\)U=U4+U123
\(\Rightarrow\)U123=U-U4=12-7,2=4,8V
mà (R1ntR2)//R3
\(\Rightarrow\)U12=U3=U123=4,8V
\(\Rightarrow\)I12=\(\frac{U_{12}}{R_{12}}=\frac{4,8}{20}=0,24A\)\(\Rightarrow\)I1=I2=I12=0,24A\(\Rightarrow\)\(\left\{{}\begin{matrix}U_1=R_1.I_1=10.0,24=2,4V\\U_2=R_2.I_2=10.0,24=2,4V\end{matrix}\right.\)
\(\Rightarrow\) I3=\(\frac{U_3}{R_3}=\frac{4,8}{5}=0,96\)A
\(=>R1nt\left(R2//R3\right)nt\left[R4//\left(R5ntR6\right)\right]\)
\(=>RTd=R1+\dfrac{R2R3}{R2+R3}+\dfrac{R4\left(R5+R6\right)}{R4+R5+R6}\)
\(=12,6+\dfrac{4.6}{4+6}+\dfrac{30\left(15+15\right)}{30+15+15}=30\left(om\right)\)
\(=>Im=\dfrac{U}{Rtd}=\dfrac{30}{30}=1A=I1=I23=I456\)
\(=>U23=I23.R23=1.\dfrac{4.6}{4+6}=2,4V=U2=U3\)
\(=>I2=\dfrac{U2}{R2}=\dfrac{2,4}{4}=0,6A=>I3=I23-I2=0,4A\)
\(=>U456=I456.R456=\dfrac{30\left(15+15\right)}{30+15+15}=15V=U4=U56\)
\(=>I4=\dfrac{U4}{R4}=\dfrac{15}{30}=0,5A\)
\(=>I56=I5=I6=\dfrac{U56}{R56}=\dfrac{15}{15+15}=0,5A\)
a) \(R_1nt(R_2//R_3)\)
\(R_1=\dfrac{U_1}{I_1}\Rightarrow U_1=0,4.14=5,6\left(V\right)\)
\(I_1=I_{AB}=0,4A\)
Có \(R_{AB}=R_1+R_{23}=14+\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{434}{19}\left(\Omega\right)\)
\(\Rightarrow U_1+U_{23}=U_{AB}=R_{AB}.I_{AB}=\dfrac{439}{19}.0,4=\dfrac{868}{95}\left(V\right)\)
\(\Rightarrow U_{23}=\dfrac{868}{95}-5,6=\dfrac{336}{95}\left(V\right)\)
\(\Rightarrow U_2=U_3=\dfrac{336}{95}\left(V\right)\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{24}{95}\left(A\right)\)
\(I_3=\dfrac{U_3}{R_3}=\dfrac{14}{95}\left(V\right)\)
b) \(U_{AB}=\dfrac{868}{95}\left(V\right)\)
\(U_{AC}=I_1.R_1=0,4.14=5,6\left(V\right)\)
\(U_{CB}=I_{23}.R_{23}=0,4.\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{336}{95}\left(V\right)\)
Vậy...
a, điện trở tưong đưong của đoạn mạch là :
\(R=\dfrac{R1.R2}{R1+R2}=\dfrac{15.10}{15+10}=6\left(\Omega\right)\)
Hiệu điện thế hai đầu đoạn mạch là : U=R.I=6.0,5=3(V)
Ta có mạch (((R5ntR6)//R4)nt(R2//R3)ntR1
R56=30\(\Omega\)=>R564=\(\dfrac{30.30}{30+30}=15\Omega\)
R23=\(\dfrac{4.6}{4+6}=2,4\Omega\)=>Rtđ=R1+R23+R456=30\(\Omega\)
=>I=I1=I23=I456=\(\dfrac{U}{Rtđ}=1A\)
Vì R2//R3=>U2=U3=U23=I23.R23=2,4V=>I2=\(\dfrac{U2}{R2}=0,6A;I3=\dfrac{U3}{R3}=0,4A\)
Vì R4//R56=>U4=U56=U456=I456.R456=15V
=>\(I4=\dfrac{U4}{R4}=0,5A\)
Vì R5ntR6=>I5=I6=I56=\(\dfrac{U56}{R56}=0,5A\)
Vậy................
hình vẽ đâu bn