A) 2.5 pF
B) 5.0 pF
C) 25 pF
D) 50 pF
Correct Answer: A
Solution :
[a] The resonant frequency \[{{f}_{r}}=\frac{1}{2\pi \sqrt{LC}}\] \[\Rightarrow \]\[{{10}^{6}}=\frac{1}{2\pi \sqrt{10\times {{10}^{-3}}\times C}}\] \[\Rightarrow \]\[C=\frac{1}{4{{\pi }^{2}}\times 10\times {{10}^{-3}}\times {{10}^{12}}}\] \[=2.5\times {{10}^{-12}}F=2.5pF\]You need to login to perform this action.
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