a) Rđ=\(\dfrac{6^2}{3}=12\Omega;Id=\dfrac{3}{6}=0,5A\)
Ta có Rtđ=Rx+R1d=4+6=10\(\Omega=>I=\dfrac{U}{Rtđ}=\dfrac{12}{10}=1,2A=>Ia=Ix=I=1,2A\)
c) Ta có Id=I1=I1d=\(1,2A=>U1=Ud=U1d=I1d.R1d=1,2.4=4,8V=>Id=\dfrac{Ud}{Id}=\dfrac{4,8}{12}=0,4A\)
=>Vì Idm=0,5> Id=0,4 => Đèn sáng yếu