$E$
$r=3\Omega$
$R_{1}=15\Omega$
$R_{2}=12\Omega$
Đèn: $6V-3W⇒R_{đ}=12\Omega$
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a, $K$ mở
$⇒E=U_{v}=6V$
b, $R_{2đ}=\dfrac{R_{2}.R_{đ}}{R_{2}+R_{đ}}=\dfrac{12.12}{12+12}=6\Omega$
$⇒R_{N}=R_{1}+R_{2đ}=15+6=21\Omega$
$⇒I=\dfrac{E}{R_{N}+r}=\dfrac{6}{21+3}=0,25A$
$⇒I_{A}=I=0,25A$
$U_{V}=U_{N}=I.R_{N}=0,25.21=5,25V$
$U_{đ}=I.R_{2đ}=0,25.6=1,5V$
$⇒$ Đèn sáng yếu hơn bình thường
