\(R_đ=\frac{U_{đm}^2}{P_{đm}}=\frac{10^2}{10}=10\Omega\)
\(\Rightarrow I_{đm}=\frac{U}{R}=\frac{10}{10}=1A\)
Ta có: \(\left(R_1//R_đ\right)ntR_2\)
\(\rightarrow R_{tđ}=\frac{R_1.R_đ}{R_1+R_đ}+R_2=\frac{40.10}{40+10}+12=8+12=20\Omega\)
\(\rightarrow I=\frac{U}{R_{tđ}}=\frac{10}{20}=0.5A\)
Do \(\left(R_1//R_đ\right)ntR_2\)
\(\Rightarrow I_{1đ}=I_2=I=0.5A\)
\(U_{1đ}=I.R_{1đ}=0.5\cdot8=4\left(V\right)\)
Do \(\left(R_1//R_đ\right)\)
\(\rightarrow U_1=U_đ=U_{1đ}=8\left(V\right)\)
\(\Rightarrow I_đ=\frac{U}{R_đ}=\frac{8}{10}=0.8\left(A\right)\)
Do \(I_đ< I_{đm}\Rightarrow\) Đèn sẽ sáng yếu hơn