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, Taco: R1= U1/I1 ; R2= U2/I2

=> I1 = U1/R1 ; I2 = U2/R2

mà I1=I2 => U1/R1 = U2/R2

=> U1/U2 = R1/R2 (1)

Ta co: Q1= U1.I1.t ; Q2 = U2.I2.t

=> \(I1=\dfrac{Q1}{U1.t};I2=\dfrac{Q2}{U2.t}\)

=> \(\dfrac{Q1}{U1.t}=\dfrac{Q2}{U2.t}\Rightarrow\dfrac{Q1}{Q2}=\dfrac{U1.t}{U2.t}\)(2)

(1) (2) => Q1/Q2 = R1/R2

b, Ta co: I1 = U1/R1 ; I2 = U2/R2

=> \(\dfrac{I1}{I2}=\dfrac{U1}{R1}:\dfrac{U2}{R2}=\dfrac{U1.R1}{U2.R2}\)

mà U1 = U2 => I1/I2 = R1/R2

\(I1=\dfrac{Q1}{U1.t};I2=\dfrac{Q2}{U2.t}\)

=> \(\dfrac{I1}{I2}=\dfrac{Q1}{U1.t}:\dfrac{Q2}{U2.t}=\dfrac{Q1}{Q2}\)

=> Q1/Q2 = R1/R2

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,\(I_1=\dfrac{U}{R_1}=\dfrac{60}{20}=3\left(\Omega\right);I_2=\dfrac{U}{R_2}=\dfrac{60}{20}=3\left(\Omega\right)\)

ta có công suất khi mắc hai điện trở // là

P1=U²(R1+R2)/R1*R2

tương tự khi mắc nối tiếp

P2=U²/(R1+R2) mà P1=4,5P2

từ hai phương trình trên ta có

(R1+R2)²=4,5*R1*R2

thay R1=4(ôm) vào ta có

giải ra ta được R2=8 hoặc R2=2

1/MCD: R1nt R2

\(R_{tđ}=R_1+R_2=15+20=35\left(\Omega\right)\)

\(I=I_2=I_1=\dfrac{U_1}{R_1}=\dfrac{6}{15}=0,4\left(A\right)\)

\(U_2=I_2\cdot R_2=0,4\cdot20=8\left(V\right)\)

\(U=I\cdot R_{tđ}=0,4\cdot35=14\left(V\right)\)

2/MCD: R1nt R2

\(R_{tđ}=R_1+R_2=6+10=16\left(\Omega\right)\)

\(I=I_1=I_2=\dfrac{U_2}{R_2}=\dfrac{5}{10}=0,5\left(A\right)\)

\(U_1=I_1\cdot R_1=0,5\cdot6=3\left(V\right)\)

\(U=I\cdot R_{tđ}=0,5\cdot16=8\left(V\right)\)