a, Khi đèn sáng bình thường, cường độ dòng điện chạy qua nó là:
$I_{đ}=\dfrac{P}{U}=\dfrac{16}{24}=\dfrac{2}{3}A$
b, Khi đèn sáng bình thường, $U_{2}=U_{đ}=24V$
$⇒U_{1}=E-U_{đ}=32-24=8V$
$⇒I_{1}=\dfrac{U_{1}}{R_{1}}=\dfrac{8}{8}=1A$
$⇒I_{2}=I_{1}-I_{đ}=1-\dfrac{2}{3}=\dfrac{1}{3}A$
c, $R_{đ}=\dfrac{U_{đ}²}{P_{đ}}=\dfrac{24²}{16}=36$ ôm
$R_{2đ}=\dfrac{R_{2}.R_{đ}}{R_{2}+R_{đ}}=\dfrac{36x}{36+x}$
$⇒R_{tđ}=R_{1}+R_{2đ}+r=8+\dfrac{36x}{36+x}+2=\dfrac{46x+360}{36+x}$
$⇒I=\dfrac{E}{R_{tđ}}=\dfrac{32}{\dfrac{46x+360}{36+x}}=\dfrac{576+16x}{23x+180}$
$⇒I_{1}=I=\dfrac{576+16x}{23x+180}$
Mà $I_{1}=\dfrac{E-U_{đ}}{R_{1}}=\dfrac{32-24}{8}=1A$
$⇒\dfrac{576+16x}{23x+180}=1$
$⇔x≈56,57$ ôm
