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Cho R1 =R2=10Ω
\(R_{12}=R_1+R_2=10+10=20\left(\Omega\right)\)
\(R_{tđ}=\dfrac{R_{12}.R_3}{R_{12}+R_3}=\dfrac{20.30}{20+30}=12\left(\Omega\right)\)
\(A=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{24}{12}=2\left(A\right)\)
\(U=U_{12}=U_3=24V\)
\(I_{12}=I_1=I_2=\dfrac{U_{12}}{R_{12}}=\dfrac{24}{20}=1,2\left(A\right)\)
\(\left\{{}\begin{matrix}U_1=I_1.R_1=1,2.10=12\left(V\right)\\U_2=I_2.R_2=1,2.10=12\left(V\right)\end{matrix}\right.\)
Vì \(R_1ntR_2\Rightarrow I_1=I_2=I_m\)
\(U=U_1+U_2=R_1\cdot I_1+R_2\cdot I_2=25\cdot I+5\cdot I=30I\left(V\right)\)
\(U_1=R_1\cdot I=15I=\dfrac{1}{2}U\)
\(U_2=R_2\cdot I=5I=\dfrac{1}{6}U\)
\(R_{tđ}=R_1+R_2=25+5=30\left(\Omega\right)\)
\(I=I_1=I_2\left(R_1ntR_2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}U=I.R_{tđ}=30I\\U_1=I_1.R_1=25I\\U_2=I_2.R_2=5I\end{matrix}\right.\)
\(\Rightarrow U>U_1>U_2\)
\(R_{tđ}=R_1+R_2=5+10=15\Omega\)
\(I_1=I_2=\dfrac{U}{R_{tđ}}=\dfrac{12}{15}=0,8A\)
\(U_1=I_1\cdot R_1=0,8\cdot5=4V\)
\(U_2=U-U_1=12-4=8V\)
\(P_1=U_1\cdot I_1=4\cdot0,8=3,2W\)
\(P_2=U_2\cdot I_2=8\cdot0,8=6,4W\)
Bài 1 :
a, TH1 : mắc nối tiếp \(R_{tđ}=R_1+R_2=30+30=60\left(\Omega\right)\)
TH2 : mắc song song \(R_{tđ}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{30.30}{60}=15\left(\Omega\right)\)
b, Vì mắc nối tiếp nên \(I_m=I_1=I_2=\dfrac{U}{R_{tđ}}=\dfrac{90}{60}=\dfrac{3}{2}\left(\Omega\right)\)
Bài 2 ;
a, \(R_{tđ}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{20.20}{40}=10\left(\Omega\right)\)
b,\(I_1=\dfrac{U}{R_1}=\dfrac{60}{20}=3\left(\Omega\right);I_2=\dfrac{U}{R_2}=\dfrac{60}{20}=3\left(\Omega\right)\)
B
Trong đoạn mạch nối tiếp công thức điện trở tương đương: \(R=R_1+R_2+...R_n\)
1,
Ta có: R\(_1\) nt R\(_2\)
\(I_1=\frac{U_1}{R_1}\) , \(I_2=\frac{U_2}{R_2}\)
Mà I\(_1\) = I\(_2\)
\(\Rightarrow\frac{U_1}{R_1}=\frac{U_2}{R_2}\)
\(\Rightarrow\frac{U_1}{U_2}=\frac{R_1}{R_2}\)
* C/m : \(R_{tđ}=R_1+R_2\)
U = U\(_1\)+U
Ta có: U\(_1\)= I.R\(_1\) , U\(_2\) = I.R\(_2\) , U=I.R\(_{tđ}\)
Mà U =U\(_1\)+U\(_2\)
=>R\(_{tđ}\)=R\(_1\)+R\(_2\)(dpcm)
* C/m \(\frac{Q_1}{Q_2}=\frac{R_1}{R_2}\)
Ta có: \(Q_1=\frac{U^2}{R_1},Q_2=\frac{U^2}{R_2}\)
\(\frac{Q_1}{Q_2}=\frac{\frac{U^2}{R_1}}{\frac{U^2}{R_2}}=\frac{R_1}{R_2}\left(đpcm\right)\)
2, Ta có: \(R_1//R_2\)
\(I_1=\frac{U_1}{R_1}\) , \(I_2=\frac{U_2}{R_2}\)
\(\rightarrow U_1=I_1.R_1\) , \(U_2=I_2.R_2\)
Mà \(U_1=U_2\)
\(\rightarrow I_1R_1=I_2R_2\)
\(\rightarrow\frac{I_1}{I_2}=\frac{R_2}{R_1}\left(đpcm\right)\)
* C/m: \(\frac{1}{R_{tđ}}=\frac{1}{R_1}+\frac{1}{R_2}\)
R\(_{tđ}\)= \(\frac{U}{I}\) = \(\frac{U}{I_1+I_2}\)
\(\rightarrow\frac{1}{R_{tđ}}=\frac{I_1+I_2}{U}\)
\(\Leftrightarrow\frac{I_1}{U}+\frac{I_2}{U}\)
\(\Leftrightarrow\frac{1}{R_1}+\frac{1}{R_2}\)
\(\rightarrow\)\(\frac{1}{R_{tđ}}=\frac{1}{R_1}+\frac{1}{R_2}\)( đpcm )
* C/m \(\frac{Q_1}{Q_2}=\frac{R_2}{R_1}\)
a) Rtd= R1+R2=5+10=15\(\left(\Omega\right)\)
b)I=I1=I2=0,3(A)
c) U1=R1.I1=5.0,3=1,5(V)
U2=R2.I2=10.0,3=3(V)
U=U1+U2=1,5+3=4,5(V)
KL......