A) \[{{\gamma }_{1}}\]
B) \[{{\gamma }_{2}}\]
C) \[\frac{{{\gamma }_{1}}-{{\gamma }_{2}}}{3}+\frac{\alpha }{3}\]
D) _ \[\frac{{{\gamma }_{1}}-{{\gamma }_{2}}}{2}+\alpha \]
Correct Answer: D
Solution :
Equivalent inductance \[{{L}_{1}}=L+2L=3L\] Equivalent inductance \[{{L}_{2}}=L\] Equivalent inductance to the circuit \[{{L}_{eq}}=\frac{{{L}_{1}}{{L}_{2}}}{{{L}_{1}}+{{L}_{2}}}=\frac{3{{L}^{2}}}{4L}=\frac{3}{4}L\] Equivalent capacitance \[{{C}_{1}}=\frac{C}{2}\] Equivalent capacitance \[{{C}_{2}}=2C\] Equivalent capacitance of the circuit \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=\frac{C}{2}+2C=\frac{5}{2}C\] Frequency of oscillation \[t=\frac{1}{2\pi }\times \frac{1}{\sqrt{{{L}_{eq}}}\sqrt{{{C}_{eq}}}}\] \[=\frac{1}{2\pi }\times \frac{1}{\sqrt{\frac{3}{4}L}}\times \frac{1}{\sqrt{\frac{5}{2}C}}=\frac{1}{2\pi }\times \frac{2}{\sqrt{3L}}\times \frac{\sqrt{2}}{\sqrt{5C}}\] \[=\frac{1}{\pi }\times \sqrt{\frac{2}{15LC}}=\frac{1}{\pi }\sqrt{\frac{2}{15}}\frac{1}{\sqrt{LC}}\]
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