a)Ta có (R1//R3)nt(R2//R4)=> Rtđ=R13+R24=\(\dfrac{R1.R3}{R1+R3}+\dfrac{R2.R4}{R2+R4}=1+2=3\Omega\)
=> I=\(\dfrac{U}{Rt\text{đ}}=\dfrac{5}{3}A\)
Vì R13ntR24=>I13=I24=I=\(\dfrac{5}{3}A\)
Vì R1//R3=> U1=U3=U13=I13.R13=\(\dfrac{5}{3}.1=\dfrac{5}{3}V\)
=> I1=\(\dfrac{U1}{R1}=\dfrac{5}{3}:2=\dfrac{5}{6}A;I3=\dfrac{U3}{R3}=\dfrac{5}{3}:2=\dfrac{5}{6}A\)
Vì R2//R4=> U2=U4=U24=I24.R24=\(\dfrac{5}{3}.2=\dfrac{10}{3}V\)
=> I2=\(\dfrac{U2}{R2}=\dfrac{10}{3}:3=\dfrac{10}{9}A;I4=\dfrac{U4}{R4}=\dfrac{10}{3}:6=\dfrac{5}{9}A\)
Vì I1<I2=> Chốt dương tại D
=> I1+Ia=I2=> Ia=I2-I1=\(\dfrac{5}{18}A\)
Vậy ampe kế chỉ 5/18 A