R_s = 3 + 2.5 + P = 5.5 + P
V = 9 V, I = 1.0 A
R_s = \(\frac{V}{I}=\frac{9}{1}= 9 \Omega\)
∴ 9 = 5.5 + P
∴ P = 9 - 5.5 = 3.5 Ω

மொத்த திறன் = 15 x 40 + 5 x 100 + 5 x 80 + 1000
= 600 + 500 + 400 + 1000
= 2500 W
P = VI, V = 220 V
\(I=\frac{P}{V}=\frac{2500}{220}=11.363 \ A\)
≃ 12 A

\(\mathrm{V}_1  =220 \mathrm{~V}, \quad \mathrm{P}_1=60 \mathrm{~W}  \)
\(\mathrm{~V}_{\mathrm{U}}  =110 \mathrm{~V}, \mathrm{P}_{\mathrm{U}}=60 \mathrm{~W}  \)
\(P  =\frac{V^2}{R} \Rightarrow R=\frac{V^2}{P}  \)
\(\therefore R_l  =\frac{V_I^2}{P_l} \text { } \ \mathrm{R}_U=\frac{V_U^2}{P_U}  \)
\(R_l  =\frac{220 \times 220}{60} \quad \mathrm{R}_{\mathrm{U}}=\frac{110 \times 110}{60}  \)
\(R_I  =\frac{48400}{60} \quad R_U=\frac{12100}{60}  \)
\(\frac{R_U}{R_l}  =\frac{12100}{60} \times \frac{60}{48400}=\frac{1}{4}  \)
\(R_U  =\frac{R_l}{4}=\frac{R}{4}\)

\(\mathrm{V}=230 \mathrm{~V}  \)
\(P=\frac{V^2}{R} \Rightarrow \mathrm{P} \propto \frac{1}{R}  \)
\(\frac{1}{R_2}=\frac{1}{\frac{R_1}{2}}+\frac{1}{\frac{R_1}{2}}=\frac{2}{R_1}+\frac{2}{R_1}=\frac{2+2}{R_1}  \)
\(\frac{1}{R_2}=\frac{4}{R_1}  \)
\(\therefore R_2=\frac{R_1}{4}  \)
\(R_1=4 R_2  \)
\(\therefore \frac{P_2}{P_1}=\frac{R_1}{R_2}=\frac{4 R_2}{R_2}=4\)

\(R \propto \frac{1}{A}, R \propto \frac{1}{r^2}  \)
\(R_A \propto \frac{1}{r_A^2}, R_B \propto \frac{1}{r_B^2} \)
\(\frac{r_A}{r_B}=\left(\frac{R_B}{R_A}\right)^{1 / 2}=\left(\frac{R_B}{3 R_B}\right)^{1 / 2}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt{3}}\)