A)
B)
C)
D)
Correct Answer: C
Solution :
In LC circuit, if \[{{\left[ \text{Fe}{{\left( {{\text{H}}_{\text{2}}}\text{O} \right)}_{\text{6}}} \right]}^{\text{3+}}}\text{}{{\left[ \text{Fe}{{\left( \text{N}{{\text{O}}_{\text{2}}} \right)}_{\text{6}}} \right]}^{\text{3-}}}\text{}{{\left[ \text{Fe}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{6}}} \right]}^{\text{3-}}}\] then \[\because \] \[{{\text{E}}_{\text{1}}}\text{=}{{\text{E}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{ }\!\!\varepsilon\!\!\text{ }}_{\text{0}}}}\text{.}\frac{\text{q}}{{{\text{r}}^{\text{2}}}}\] so \[{{E}_{R}}=\sqrt{E_{1}^{2}+E_{2}^{2}+2{{E}_{1}}{{E}_{2}}\cos {{60}^{\circ }}}\] As \[=\sqrt{E_{1}^{2}+E_{1}^{2}+2E_{1}^{2}\times \frac{1}{2}=\sqrt{3}{{E}_{1}}}\] is the natural frequency of LC circuit, therefore for an LC circuit if the frequency of applied AC becomes equal to the natural frequency of an AC circuit then the amplitude of current becomes infinite due to zero impedance. Hence, the correct diagram is option (c).
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