Cách 1 :

Ta có:

\(\begin{array}{l}{I_1} = {\omega _1}q = \frac{q}{{\sqrt {{L_1}C} }} = > {L_1} = \frac{{{q^2}}}{{{I_1}^2C}}\\{I_2} = {\omega _2}q = \frac{q}{{\sqrt {{L_2}C} }} = > {L_2} = \frac{{{q^2}}}{{{I_2}^2C}}\end{array}\)

=> Khi sử dụng cuộn cảm có độ tự cảm \({L_3} = 13{L_1} + 9{L_2}\) thì cường độ dòng điện cực đại là :

\({I_3} = \frac{q}{{\sqrt {{L_3}C} }} = \frac{q}{{\sqrt {(13\frac{{{q^2}}}{{{I_1}^2C}} + 9\frac{{{q^2}}}{{{I_2}^2C}})C} }} = \frac{{{I_1}{I_2}}}{{\sqrt {13{I_2}^2 + 9{I_1}^2} }} = 10mA\)

Cách 2:

Ta có:

\(\begin{array}{l}{I_1} = {\omega _1}q = \frac{q}{{\sqrt {{L_1}C} }};{I_2} = {\omega _2}q = \frac{q}{{\sqrt {{L_2}C} }}\\ \to \frac{{{I_1}}}{{{I_2}}} = \sqrt {\frac{{{L_2}}}{{{L_1}}}} = \frac{{70}}{{35}} \to \frac{{{L_2}}}{{{L_1}}} = 4\\ \to {L_2} = 4{L_1}\end{array}\)

\(\begin{array}{l}{I_3} = \frac{q}{{\sqrt {{L_3}C} }} \to \frac{{{I_3}}}{{{I_1}}} = \sqrt {\frac{{{L_1}}}{{{L_3}}}} = \sqrt {\frac{{{L_1}}}{{13{L_1} + 9{L_2}}}} = \sqrt {\frac{{{L_1}}}{{13{L_1} + 9.4{L_1}}}} = \frac{1}{7}\\ \to {I_3} = \frac{{{I_1}}}{7} = \frac{{70}}{7} = 10mA\end{array}\)