Tiếp nè
Do \(R_{12}//R_3\)
\(\Rightarrow U_{12}=U_3=U_{123}=\frac{48a}{7}\left(V\right)\)
\(\rightarrow I_3=\frac{U_3}{R_3}=\frac{\frac{48a}{7}}{4a}=\frac{12}{7}\left(A\right)\)
\(\rightarrow I_{12}=\frac{U_{12}}{R_{12}}=\frac{\frac{48a}{7}}{3a}=\frac{16}{7}\left(A\right)\)
Do \(R_1ntR_2\)
\(\Rightarrow I_1=I_2=I_{12}=\frac{16}{7}\left(A\right)\)
Vậy:
\(I_4=4\left(A\right)\)
\(I_3=\frac{12}{7}\left(A\right)\)
\(I_1=I_2=\frac{16}{7}\left(A\right)\)
Ta có : [(R\(_1\) nt R\(_2\) ) \(//\) R\(_3\) ] nt R\(_4\)
\(\Rightarrow\) R\(_{12}\) = R\(_1\) + R\(_2\) = a+2a = 3a (\(\Omega\) )
\(\Rightarrow R_{234}=\frac{R_{12}.R_3}{R_{12}+R_3}=\frac{3a.4a}{3a+4a}=\frac{12a^2}{7a}=\frac{12a}{7}\left(\Omega\right)\)
\(\Rightarrow R_{td}=R_{123}+R_4=\frac{12a}{7}+a=\frac{19a}{7}\left(\Omega\right)\)
Do \(R_{123}ntR_4\)
\(\Rightarrow I_{123}=I_4=I=4\left(A\right)\)
\(\Rightarrow U_{123}=I_{123}.R_{123}=\frac{12a}{7}.4=\frac{48a}{7}\left(V\right)\)
Do
