a, \(=>R4nt\left[R1//\left(R2ntR3\right)\right]\)
\(=>Rtd=R4+\dfrac{R1\left(R2+R3\right)}{R1+R2+R3}=2+\dfrac{6\left(6+6\right)}{6+6+6}=6\left(om\right)\)
\(=>Im=\dfrac{U}{Rtd}=\dfrac{18}{6}=3A=I4=I123\)
\(=>U4=I4.R4=3.2=6V\)
\(=>U123=I123.R123=3.\dfrac{6\left(6+6\right)}{6+6+6}=12V=U23\)
\(=>I23=\dfrac{U23}{R23}=\dfrac{12}{6+6}=1A=I3=>U3=I3.R3=6V\)
\(=>Uv=U4+U3=6+6=12V\)
b, \(=>R2//\left[R1nt\left(R3//R4\right)\right]\)
\(=>U2=U134=Uab=18V\)
\(=>I2=\dfrac{U2}{R2}=\dfrac{18}{6}=3A\)
\(=>I134=\dfrac{U134}{R134}=\dfrac{18}{6+\dfrac{2.6}{2+6}}=2,4A=I34\)
\(=>U34=I34.R34=2,4.\dfrac{2.6}{2+6}=3,6V=U3\)
\(=>I3=\dfrac{U3}{R3}=\dfrac{3,6}{6}=0,6A\)
Giả sử chiều dòng điện qua ampe kế từ M tới B
\(=>Ia=I3+I2=0,6+3=3,6A\)
do đó số chỉ ampe kế là 3,6A
