Question

Tóm tắt 

I\(_2\) = 3A

R\(_1\) = 12\(\Omega\)

R\(_2\) = 10\(\Omega\)

R\(_3\) = 15\(\Omega\)

a) R\(_{tđ}\) = ?\(\Omega\)

b) U = ?V

Answer 1

\(R_1//R_{23}\Rightarrow R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{10\cdot15}{10+15}=6\Omega\)

\(R_1ntR_{23}\Rightarrow R_{tđ}=R_1+R_{23}=12+6=18\Omega\)

\(U=I\cdot R=3\cdot18=54V\)

Answer 2

có: \(R_1nt\left(R_2//R_3\right)\)

=> \(R_{tđ}=R_1+\dfrac{R_2.R_3}{R_2+R_3}=12+\dfrac{10.15}{10+15}=18\Omega\)

có: \(R_2//R_3\)

=> \(\dfrac{R_2}{R_3}=\dfrac{I_3}{I_2}\Leftrightarrow I_3=\dfrac{10}{15}.3=2A\)

Từ trên suy ra được: 

\(I_1=I_2+I_3=3+2=5A\)

=> \(U=I.R=5.18-90V\)