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A constant voltage at a frequency of `1 MHz` is applied to an inductor in series with variable capacitor, when capacitor is `500 pF`, the current has its maximum value, while it is reduced to half when capacitance is `600 pF`. Find `Q` factor of the circuit isA. 10.4B. 20.8C. 5.2D. 9.4 |
Answer» Correct Answer - A `Q = (omega_(0) L)/(R ) = (314)/(30.01) = 10.4` |
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