Question

Answer 1

a, Vì R₂ và R₃ mắc nt nên R₂₃=R₂+R₃=10Ω

Vì R₂₃//R₁ nên \(\dfrac{1}{R_{td}}=\dfrac{1}{R_{23}}+\dfrac{1}{R_1}=\dfrac{2}{10}=\dfrac{1}{5}\Rightarrow R_{td}=5\text{Ω}\)

b, Ta có \(U=R_{td}\cdot I=7,5\left(V\right)\)

Mà R₁//R₂₃ nên U=U₁=U₂₃=7,5(V)

\(\Rightarrow I_1=\dfrac{U_1}{R_1}=0,75A;I_{23}=\dfrac{U_{23}}{R_{23}}=0,75A\)

Mà R₂ nt R₃ nên \(I_{23}=I_2=I_3=0,75A\)

Vậy \(I_1=I_2=I_3=0,75A\)

 

Answer 2

\(R_1nt\left(R_2//R_3\right)\)

a) Điện trở tương đương mạch:

   \(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{4\cdot6}{4+6}=2,4\Omega\)

 \(R=R_1+\dfrac{R_2\cdot R_3}{R_2+R_3}=10+\dfrac{4\cdot6}{4+6}=12,4\Omega\)

b) \(U_m=I_m\cdot R=1,5\cdot12,4=18,6V\)

     \(\Rightarrow U_1=U_{23}=18,6V\)

     \(\Rightarrow I_1=\dfrac{U_1}{R_1}=\dfrac{18,6}{10}=1,86A\)

        \(I_{23}=\dfrac{U_{23}}{R_{23}}=7,75A\)\(\Rightarrow I_2=I_3=I_{23}=7,75A\)