Để đèn 1 sáng bình thường tức \(I_{đm1}=\dfrac{P_1}{U_1}=\dfrac{60}{60}=1A\)
\(\Rightarrow I_{Đ1+Đ2}=I_{đm1}=1A\)
\(R_{Đ1}=\dfrac{U_1^2}{P_1}=\dfrac{60^2}{60}=60\Omega\)\(;R_2=\dfrac{U_{Đ2}^2}{P_2}=\dfrac{60^2}{120}=30\Omega\)
\(\Rightarrow R_{Đ1+Đ2}=60+30=90\Omega\Rightarrow U_2=U_{Đ1+Đ2}=1\cdot\left(60+30\right)=90V\)
\(R_{2+Đ1+Đ2}=\dfrac{90\cdot15}{90+15}=\dfrac{90}{7}\Omega\)
\(\Rightarrow I_1=I_2=\dfrac{90}{\dfrac{90}{7}}=7A\)\(\Rightarrow U_1=I_1R_1=7\cdot15=105V\)
Vậy \(U_{AB}=U_1+U_2=105+90=195V\)