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

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 \(\frac{Q_1}{Q_2}=\frac{R_2}{R_1}\)

trong đoạn mạch R1ntR2 

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

nếu nt mà R1=R2 với I1=I2 thì mới có TH nt có hệ \(\dfrac{U_1}{U_2}=\dfrac{I_1}{I_2}\)

- Trong đoạn mạch mắc song song ta có: U₁=U₂=U_m và Im=I₁+I₂

và R_m=\(\dfrac{R_1.R_2}{R_1+R_2}\)

I₁=\(\dfrac{U_1}{R_1}=\dfrac{U_M}{R_1}=\dfrac{I_m.R_m}{R_1}=\dfrac{R_m}{R_1}.I_m=\dfrac{\dfrac{R_1.R_2}{R_1+R_2}}{R_1}.I_m=\dfrac{R_2}{R_1+R_2}.I_m\)

\(I_1\cdot R_1=I_2\cdot R_2\)

ta có sơ đồ:

Ta có: R₁₂=\(\dfrac{R_1R_2}{R_1+R_2}=\dfrac{10.20}{10+20}=\dfrac{200}{30}=\dfrac{20}{3}\left(\Omega\right)\)

R₁₂₃=R₁₂+R₃=\(\dfrac{20}{3}+30=\dfrac{110}{3}\left(\Omega\right)\)

=> R_td=R₁₂₃₄=\(\dfrac{R_{123}R_4}{R_{123}+R_4}=\dfrac{\dfrac{110}{3}.40}{\dfrac{110}{3}+40}=\dfrac{440}{23}=19,13\left(\Omega\right)\)

=> I=\(\dfrac{U}{R_{td}}=\dfrac{90}{\dfrac{440}{23}}=\dfrac{207}{44}=4,7\left(A\right)\)

Lại có:

U=U₄=U₁₂₃=90(V)

=> I₄=U₄:R₄=90:40=2,25(A)

I₁₂=I₃=U₁₂₃:R₁₂₃=\(\dfrac{90}{\dfrac{110}{3}}=2,45\left(A\right)\)

U₁₂=U₁=U₂=U-U₃=U-I₃R₃=90-\(\dfrac{27}{11}.30\)=\(\dfrac{180}{11}=16,36\left(V\right)\)

=> I₁=\(\dfrac{U_1}{R_1}=\dfrac{\dfrac{180}{11}}{10}=\dfrac{18}{11}=1,636\left(A\right)\)

I₂\(=\dfrac{U_2}{R_2}=\dfrac{\dfrac{180}{11}}{20}=\dfrac{9}{11}=0,818\left(A\right)\)

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

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Đáp án D

Giữa I 1 ,   I 2 ,   I 3  có mối liên hệ là I 2   =   I 3   =   I 1 / 2