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A capacitor of capacitance `2 mu F` is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is `2 mA`, the voltage across the capacitor will beA. `0.16 V`B. `0.32 V`C. `79.5 V`D. `159 V` |
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Answer» Correct Answer - A (a) Reactance of capacitor or capactive reactance is given by `X_(C ) = (1)/(omega C) = (1)/(2 pi fC)` where `f` is frequency of the alternating current Given, `f = 1 xx 10^(3) Hz, C = 2 xx 10^(-6) F` `implies X_(C ) = (1)/(2 xx 3.14 xx 10^(3) xx 2 xx 10^(-6))` `= (10^(3))/(4 xx 3.14)` Hence, voltage across the capacitor is `V = i X_(C ) = 2 xx 10^(-3) xx (10^(3))/(4 xx 3.14) = 0.16 V` |
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