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A capacitor of 250 mu F is connected parallel with a inductor of 0.16mH. IF the effective resistanceis 20 Omega then resonant frequency……. Hz. |
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Answer» `9 times 10^4` `THEREFORE f_0=1/(2 times 3.14 times sqrt(0.16 times 10^-3 times 250 times 10^-6))` `therefore f_0=1/(6.28 times sqrt(16 times 10^-5 times 25 times 10^-5))` `therefore f_0=1/(6.28 times sqrt(400 times 10^-10))` `therefore f_0=1/(6.28 times 20 times 10^-5)` `therefore f_0=1/(6.28 times 20) times 10^5` `therefore f_0=7.96 times 10^-3 times 10^5` `therefore f_0=7.96 times 10^2 Hz` |
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