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bài 1 ( nhx R nào mình ko nhắc đến thì có nghĩa nó ko có cđ dđ qua bn nhé)
a, mạch vẽ lại R2ntR1
\(R_{tđ}=2+6=8\left(\Omega\right)\)
\(I_1=I_2=\dfrac{9}{8}=1,125\left(A\right)\)
b, (R1ntR2)//R5
\(R_{tđ}=\dfrac{8.10}{18}=\dfrac{40}{9}\left(\Omega\right)\)
\(I_1=I_2=\dfrac{9}{8}=1,125\left(A\right)\)
\(I_5=\dfrac{9}{10}=0,9\left(A\right)\)
c, R2nt[(R3ntR4)//R1]
\(R_{tđ}=6+\dfrac{2.14}{16}=7,75\left(\Omega\right)\)
\(I_2=\dfrac{9}{7,75}=\dfrac{36}{31}\left(A\right)\)
\(U_{134}=9-\dfrac{36}{31}.6\approx2\left(V\right)\)
\(I_3=I_4=\dfrac{2}{14}=\dfrac{1}{7}\left(A\right)\)
\(I_1=\dfrac{2}{2}=1\left(A\right)\)
d, mạnh như hình
\(R_{AB}=7,75\left(\Omega\right)\)
\(R_{tđ}=\dfrac{10.7,75}{17,75}=\dfrac{310}{71}\)
I1 I2 I3 I4 như ý c
\(I_5=\dfrac{9}{10}=0,9\left(A\right)\)
\(R1ntR2ntR3\)
\(a,\Rightarrow R3=Rtd-R1-R2=20\Omega\)
\(b,\Rightarrow I1=I2=I3=1,2A\)
\(c,\Rightarrow Um=ImRtd=100.1,2=120V\)
\(\Rightarrow\left\{{}\begin{matrix}U1=I1.R1=54V\\U2=I2R2=42V\\U3=I3R3=24V\end{matrix}\right.\)
R1 n t (R2//R3//R4)
a,\(=>\dfrac{1}{R234}=\dfrac{1}{R2}+\dfrac{1}{R3}+\dfrac{1}{R4}=>R234=10\left(om\right)\)
\(=>Rmp=R1+R234=25\left(ôm\right)\)
b
ta thấy R2=R3=R4 mà U2=U3=U4
=>I2=I3=I4=0,5A
\(=>I1=I2+I3+I4=1,5A\)
c,\(U2=U3=U4=I2.R2=15V\)
\(U1=I1.R1=22,5V=>Ump=U1+U2=37,5V\)
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)\(I_1=I_{234}=I_{AB}=\dfrac{U_{AB}}{R_{AB}}=\dfrac{30}{\dfrac{201}{23}}=\dfrac{230}{67}A\approx3,4A\)
\(U_{23}=U_4=U-U_1=30-I_1\cdot R_1=30-\dfrac{230}{67}\cdot3=\dfrac{1320}{67}V\)
\(I_4=\dfrac{U_4}{R_4}=\dfrac{\dfrac{1320}{67}}{11}=\dfrac{120}{67}A\approx1,79A\)
\(I_2=I_3=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{\dfrac{1320}{67}}{12}=\dfrac{110}{67}A\approx1,64A\)
,R2=R3=R4=20 
đoạn mạch nối tiếp hay song song 
R3=30:4=7,5\(\Omega\)
Ta có R4nt(R1//R2//R3)
=>Rtđ=R1+\(\dfrac{R1.R2.R3}{\left(R1.R2\right)+\left(R2.R3\right)+\left(R3.R1\right)}=25+\dfrac{10}{3}=\dfrac{250}{3\Omega}\)(\(\dfrac{250}{3}\Omega nh\text{é}\left(\right)\))
b)I=\(\dfrac{U}{Rt\text{đ}}=45:\dfrac{250}{3}=0,54A\)
Vì R4ntR123=>I4=I123=I=0,54A
Vì R1//R2//R3=>U1=U2=U3=U123=I123.R123=0,54.\(\dfrac{10}{3}=1,8V\)
=>I1=\(\dfrac{U1}{R1}=\dfrac{1,8}{30}=0,06A;I2=\dfrac{U2}{R2}=\dfrac{1,8}{30}=0,06A;I3=\dfrac{U3}{R3}=\dfrac{1,8}{7,5}=0,24A\)