Câu hỏi 3

Ta có

\(\eqalign{
& {I_1} = {I_2} = {{{I_{max}}} \over n} = > {Z_1} = {Z_2} = > {\omega _1}L - {1 \over {{\omega _1}C}} = - {\omega _2}L + {1 \over {{\omega _2}C}} \cr
& \omega _0^2 = {\omega _1}{\omega _2} = {1 \over {LC}} = > {\omega _2}L = {1 \over {{\omega _1}C}};{I_1} = {{{I_{max}}} \over n} \cr
& = > {U \over {\sqrt {{R^2} + {{\left( {{\omega _1}L - {1 \over {{\omega _1}C}}} \right)}^2}} }} = {1 \over n}{U \over R} = > {n^2}{R^2} = {R^2} + {\left( {{\omega _1}L - {1 \over {{\omega _1}C}}} \right)^2} = {R^2} + {\left( {{\omega _1}L - {\omega _2}L} \right)^2} \cr
& = > \left( {{n^2} - 1} \right){R^2} = {\left( {{\omega _1} - {\omega _2}} \right)^2}{L^2} = > R = {{L\left( {{\omega _1} - {\omega _2}} \right)} \over {\sqrt {\left( {{n^2} - 1} \right)} }} \cr} \)