a;
K mở \(=>R3nt\left[\text{R4/}/\left(R1ntR2\right)\right]\)
\(=>Rtd=R3+\dfrac{R4\left(R1+R2\right)}{R4+R1+R2}=4+\dfrac{6\left(8+4\right)}{6+8+4}=8\left(om\right)\)
\(=>Im=Ia=\dfrac{6}{8}=0,75A\)
K đóng \(=>R1//\left[R4nt\left(R2//R3\right)\right]\)
\(=>Rtd=\dfrac{8.\left[6+\dfrac{4.4}{4+4}\right]}{8+6+\dfrac{4.4}{4+4}}=4\left(om\right)\)
\(=>U1234=Uab=U234=6V\)
\(=>I234=\dfrac{6}{6.+\dfrac{4.4}{4+4}}=\dfrac{6}{8}=0,75A=I23\)
\(=>U23=U3=0,75.\dfrac{4.4}{4+4}=1,5A=Ia\)
b, theo đề ra \(=>Ia2=0A\)
=>bỏ điện trở R2 mạch đưa về dạng \(\left(R1ntR5\right)//\left(R4ntR3\right)\)
mạch cân bằng \(=>\dfrac{R1}{R4}=\dfrac{R5}{R3}=>\dfrac{8}{6}=\dfrac{R5}{4}=>R5=5,3\left(om\right)\)
