Hướng dẫn: Dùng chuẩn hóa.

ω (rad/s)	*Z_L* (chuẩn hóa)	Z_C	Công suất: \[P = \frac{{{U^2}.R}}{{{R^2} + {{({Z_L} - {Z_C})}^2}}}\]	Hệ số công suất: \[\user2{cos\varphi = }\frac{\user2{R}}{{\sqrt {{\user2{R}^\user2{2}}\user2{ + (}{\user2{Z}_L}\user2{ - Z}_C^{}{\user2{)}^2}} }}\]
50π	1	x	\[{P_1} = \frac{{{U^2}.R}}{{{R^2} + {{(1 - x)}^2}}}.\]	\[\user2{cos}{\user2{\varphi }_1}\user2{ = }\frac{\user2{R}}{{\sqrt {{\user2{R}^\user2{2}}\user2{ + (1 - x}{\user2{)}^2}} }}\]
100π	2	x/2	\[{P_2} = \frac{{{{(2U)}^2}.R}}{{{R^2} + {{(2 - \frac{x}{2})}^2}}}.\]	\[\user2{cos}{\user2{\varphi }_2}\user2{ = }\frac{\user2{R}}{{\sqrt {{\user2{R}^\user2{2}}\user2{ + (2 - x/2}{\user2{)}^2}} }}\]
150π	3	x/3	\[{P_3} = \frac{{{{(3U)}^2}.R}}{{{R^2} + {{(3 - \frac{x}{3})}^2}}}.\]	\[\user2{cos}{\user2{\varphi }_3}\user2{ = }\frac{\user2{R}}{{\sqrt {{\user2{R}^\user2{2}}\user2{ + (3 - x/3}{\user2{)}^2}} }}\]

Khi: ω =50π rad/s: \[{P_1} = \frac{{{U^2}.R}}{{{R^2} + {{({Z_{L1}} - {Z_{C1}})}^2}}} = \frac{{{U^2}.R}}{{{R^2} + {{(1 - x)}^2}}}\];

Khi: ω =100π rad/s: \[{P_2} = \frac{{{{(2U)}^2}.R}}{{{R^2} + {{({Z_{L2}} - {Z_{C2}})}^2}}} = \frac{{{{(2U)}^2}.R}}{{{R^2} + {{(2 - \frac{x}{2})}^2}}} = 4{P_1} = 4\frac{{{U^2}.R}}{{{R^2} + {{(1 - x)}^2}}}.\]

=> \[\begin{array}{l}{(1 - x)^2} = {(2 - \frac{x}{2})^2} \Leftrightarrow 1 - 2{\rm{x}} + {x^2} = 4 - 2{\rm{x + }}\frac{{{x^2}}}{4} = > \frac{{3{x^2}}}{4} = 3 = > x = 2.\\\end{array}\]

Khi: ω =150π rad/s: \[{P_3} = \frac{{{{(3U)}^2}.R}}{{{R^2} + {{({Z_{L3}} - {Z_{C3}})}^2}}} = \frac{{{{(3U)}^2}.R}}{{{R^2} + {{(3 - \frac{x}{3})}^2}}} = \frac{{81}}{{13}}{P_1} = \frac{{81}}{{13}}.\frac{{{U^2}.R}}{{{R^2} + {{(1 - x)}^2}}}.\]

\[\begin{array}{l}13{R^2} + 13{(1 - x)^2} = 9({R^2} + {(3 - \frac{x}{3})^2}) \Rightarrow 4{R^2} + 13 - 26{\rm{x}} + 13{x^2} = 81 - 18{\rm{x + }}{x^2}\\ \Leftrightarrow 4{R^2} = - 12{x^2} + 8x + 68 = > {R^2} = - 3{x^2} + 2x + 17\\x = 2 = > R = 3\end{array}\]

\[{\user2{k}_\user2{1}}\user2{ = cos}{\user2{\varphi }_1}\user2{ = }\frac{\user2{R}}{{\sqrt {{\user2{R}^\user2{2}}\user2{ + (1 - x}{\user2{)}^2}} }}\user2{ = }\frac{3}{{\sqrt {{3^\user2{2}}\user2{ + (1 - 2}{\user2{)}^\user2{2}}} }}\user2{ = }\frac{3}{{\sqrt {10} }}\user2{ = 0,94886}\user2{.}\]

\[{\user2{k}_3}\user2{ = cos}{\user2{\varphi }_3}\user2{ = }\frac{\user2{R}}{{\sqrt {{\user2{R}^\user2{2}}\user2{ + (3 - x/3}{\user2{)}^2}} }}\user2{ = }\frac{3}{{\sqrt {{3^\user2{2}}\user2{ + (3 - 2/3}{\user2{)}^\user2{2}}} }}\user2{ = 0,78935}\user2{.}\]Chọn B.

Cường độ dòng điện i = 2cos (100pit) (A) có giá trị cực đại là A. 2A B. căn bậc hai A