a)\(R_1//R_2\Rightarrow U_1=U_2=U=220V\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{220}{40}=5,5A;I_2=\dfrac{U_2}{R_2}=\dfrac{220}{60}=\dfrac{11}{3}A\)
b)Tiết diện sợi dây: \(R=\rho\cdot\dfrac{l}{S}\)
\(\Rightarrow S=\rho\cdot\dfrac{l}{R_1}=0,4\cdot10^{-6}\cdot\dfrac{12000}{40}=1,2\cdot10^{-4}m^2=1,2cm^2\)
c)\(R_Đ=\dfrac{U^2_Đ}{P_Đ}=\dfrac{220^2}{60}=\dfrac{2420}{3}\Omega;I_{Đđm}=\dfrac{P_Đ}{U_Đ}=\dfrac{60}{220}=\dfrac{3}{11}\approx0,273A\)
\(\left(R_1//R_2\right)ntR_Đ\)
\(R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{40\cdot60}{40+60}=24\Omega\)
\(R_{tđ}=R_{12}+R_Đ=24+\dfrac{2420}{3}=\dfrac{2492}{3}\Omega\)
\(I_Đ=I_{mạch}=\dfrac{U}{R_{tđ}}=\dfrac{220}{\dfrac{2492}{3}}=\dfrac{165}{623}\approx0,265A< I_{Đđm}\)
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