Question

Giúp mình với mình đang kiểm tra

Answer 1

Bài 1.

\(R_{Đ1}=\dfrac{U^2_{Đ1}}{P_{Đ1}}=\dfrac{6^2}{3}=12\Omega\)

\(R_2=\dfrac{U^2_{Đ2}}{P_{Đ2}}=\dfrac{6^2}{6}=6\Omega\)

\(I_{Đ1đm}=\dfrac{P_{Đ1}}{U_{Đ1}}=\dfrac{3}{6}=0,5A\)

\(I_{Đ2đm}=\dfrac{P_{Đ2}}{U_{Đ2}}=\dfrac{6}{6}=1A\)

a)Khi K mở: \(Đ_2ntĐ_1\)

   \(R_{tđ}=R_{Đ1}+R_{Đ2}=12+6=18\Omega\)

   \(I_{Đ1}=I_{Đ2}=I_m=\dfrac{U}{R}=\dfrac{12}{18}=\dfrac{2}{3}A\)

   \(U_{Đ1}=\dfrac{2}{3}\cdot12=8V;U_{Đ2}=\dfrac{2}{3}\cdot6=4V\)

   Vậy đèn 2 sáng hơn đèn 1.