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\(12.R1//R2\Rightarrow\left\{{}\begin{matrix}a,\Rightarrow U=U1=I1.R1=20.4=80V\\\Rightarrow R2=\dfrac{U}{I2}=\dfrac{80}{2,2}=\dfrac{400}{11}\left(\Omega\right)\\b,R2//R3\Rightarrow\dfrac{R2.R3}{R2+R3}=\dfrac{U}{I'}=\dfrac{80}{5,2}=\dfrac{200}{13}\Rightarrow R3\approx26,67\left(\Omega\right)\\\Rightarrow I2=I'-I3=5,2-\dfrac{U}{R3}\approx2,2A\end{matrix}\right.\)

\(13\Rightarrow\left\{{}\begin{matrix}R1ntR2\Rightarrow Im=\dfrac{U}{R1+R2}\Rightarrow\dfrac{90}{R1+R2}=1\\R1//R2\Rightarrow Im=\dfrac{U}{\dfrac{R1.R2}{R1+R2}}=\dfrac{90\left(R1+R2\right)}{R1.R2}=4,5\\\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}R1+R2=90\\90\left(R1+R2\right)=4,5.R1R2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}R2=90-R1\\90\left(R1+90-R1\right)=4,5.R1\left(90-R1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}R2=90-60=30\Omega\\R2=90-30=60\Omega\end{matrix}\right.\\\left[{}\begin{matrix}R1=60\Omega\\R2=30\Omega\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left(R1;R2\right)=\left\{\left(30;60\right);\left(60;30\right)\right\}\)

\(14.\Rightarrow\left\{{}\begin{matrix}a,\Rightarrow Rtd=\dfrac{R1R2}{R1+R2}=\dfrac{2R2^2}{3R2}=\dfrac{U}{I}=\dfrac{48}{2}=24\Rightarrow\left\{{}\begin{matrix}R2=36\Omega\\R1=2.R2=72\Omega\end{matrix}\right.\\b,R1ntR2\Rightarrow U=I\left(R1+R2\right)=2\left(36+72\right)=216V\\\\\end{matrix}\right.\)

\(15.\Rightarrow\dfrac{1}{Rtd}=\dfrac{1}{\dfrac{U}{I}}=\dfrac{1}{\dfrac{60}{9}}=\dfrac{3}{20}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}=\dfrac{1}{R1}+\dfrac{1}{\dfrac{R1}{2}}+\dfrac{1}{\dfrac{R1}{3}}\Rightarrow\left\{{}\begin{matrix}R1=40\Omega\\R2=\dfrac{R1}{2}=20\Omega\\R3=\dfrac{R1}{3}=\dfrac{40}{3}\Omega\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}I1=\dfrac{U}{R1}=\dfrac{60}{40}=1,5A\\I2=\dfrac{U}{R2}=\dfrac{60}{20}=3A\\I3=\dfrac{U}{R3}=\dfrac{60}{\dfrac{40}{3}}=4,5A\end{matrix}\right.\)

Tóm tắt

R1= 15 ôm

R2=25 ôm

R3 = 40 ôm

U2 = 10 V

a, Rtđ = ?

b, U3 =?

a, Rtđ = R1 + R2 +R3= 15 + 25 +40 = 80 Ω

b,ADCT \(\dfrac{U2}{U3}=\dfrac{R2}{R3}\)

T/s \(\dfrac{10}{U3}=\dfrac{25}{40}\)

=> U3 = 10.40/25 = 16 ( V)

\(=>R1ntR2ntR3=>Rtd=R1+R2+R3=3R1+R2\left(om\right)\)

\(=>RTd=\dfrac{12}{0,5}=24\left(om\right)\)

\(=>3R1+R2=24=>R2=24-3R1\)

\(I=I1=I2=I3=0,5A\)

\(=>3U1=U2\)\(=>3.0,5.R1=R2.0,5=>3R1=R2=>3R1=24-3R1=>R1=4\left(om\right)\)

\(=>R2=24-3R1=12\left(om\right)\)

\(=>R3=2R1=8\left(om\right)\)

\(=>U1=0,5.R1=2V\)

\(=>U2=0,5.R2=6V\)

\(=>U3=0,5.8=4V\)

Cho ba điện trở R1 = R2 = 10 , R3 = 20 . R1 mắc song R2, R1 và R2 mắc nối tiếp với R3. Điện trở tương đương của đoạn mạch là: A. 10Ω B.15Ω C.20Ω D.25Ω

Giải thích:

\(R_3nt\left(R_1//R_2\right)\)

\(R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{10\cdot10}{10+10}=5\Omega\)

\(R_{tđ}=R_3+R_{12}=20+5=25\Omega\)

Chọn D.

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...

Bài 2 :

Tóm tắt :

\(R_1=R_2=R_3=40\Omega\)

\(U_{AB}=10V\)

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\(R_{tđ}=?;I=?;I_1=?I_2=?I_3=?\)

\(U_1=?;U_2=?;U_3=?\)

TH₁ : \(R_1//\left(R_2ntR_3\right)\)

TH₂ : \(R_2nt\left(R_3//R_1\right)\)

TH3 : R1 //R2//R3

GIẢI :

Trường hợp A :

R1 R2 R3 + - R1//(R2nối tiếp R3)

Điện trở tương đương toàn mạch là :

\(R_{tđ}=\dfrac{R_1.R_{23}}{R_1+R_{23}}=\dfrac{40.\left(40+40\right)}{40+80}\approx26,67\left(\Omega\right)\)

Cường độ đòng điện I là :

\(I=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{10}{26,67}\approx0,37\left(A\right)\)

Vì R1//R23 => \(U_{AB}=U_1=U_{23}=10V\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{10}{40}=0,25\left(A\right)\)

\(I=I_1+ I_{23}\Rightarrow I_{23}=I-I_1=0,37-0,25=0,12\left(A\right)\)

Vì R2 ntR3 => \(I_2=I_3=I_{23}=0,12A\)

\(\left\{{}\begin{matrix}U_2=I_2.R_2=0,12.40=4,8\left(V\right)\\U_3=U_2=4,8\left(V\right)\end{matrix}\right.\)

Trường hợp B :

R2 R3 R1 A B

Vì R2 nt(R3//R1) nên :

\(R_{tđ}=R_2+\dfrac{R_3.R_1}{R_3+R_1}=40+\dfrac{40.40}{40+40}=60\left(\Omega\right)\)

Cường độ dòng điện I là :

\(I=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{10}{60}=\dfrac{1}{6}\left(A\right)\)

=> \(I=I_2=I_{31}=\dfrac{1}{6}\left(A\right)\)

\(U_2=I_2.R_2=\dfrac{1}{6}.40\approx6,67\left(V\right)\)

\(U_{31}=U_{AB}-U_2=3,33\left(V\right)\)

Mà : R3//R1 => \(U_{31}=U_3=U_1=3,33V\)

\(\left\{{}\begin{matrix}I_3=\dfrac{U_3}{R_3}=\dfrac{3,33}{40}=0,08325\left(A\right)\\I_1=I_3=0,08325\left(A\right)\end{matrix}\right.\)

Trường hợp C :

R1 R2 R3 + -

Vì R1//R2//R3 nên :

Điện trở tương đương toàn mạch là :

\(R_{tđ}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}=\dfrac{1}{\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}}=\dfrac{40}{3}\left(\Omega\right)\)

\(U_{AB}=U_1=U_2=U_3=10V\)

Cường độ dòng điện I là :

\(I=\dfrac{U}{R_{tđ}}=\dfrac{10}{\dfrac{40}{3}}=0,75\left(A\right)\)

\(I_1=I_2=I_3=\dfrac{U_1}{R_1}=\dfrac{10}{40}=0,25\left(A\right)\)