ampe kế lí tưởng ta vẽ lại mạch có
(R1//R2//R3)nt R4
ta có R1=12
R2=6
R3=R4=4
\(\frac{1}{R_{123}}=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}\Rightarrow R_{123}=2\left(\Omega\right)\)
\(R_{td}=R_4+R_{123}=2+4=6\Rightarrow I=\frac{12}{6}=2\left(A\right)\)(IA=2A)
\(U_{123}=12-U_4=12-\left(2.4\right)=4\left(V\right)\)
\(\Rightarrow I_1=\frac{4}{12}=\frac{1}{3}\left(A\right)\)
mà \(I-I_1=I_{A1}=2-\frac{1}{3}=\frac{5}{3}\left(A\right)\)
\(I_3=\frac{4}{4}=1\left(A\right)\)\(\Rightarrow I_{A2}=I-I_3=2-1=1\left(A\right)\)