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if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a de circuit after the steady state. |
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Answer» Solution :`I_(max)= (E_(max) )/(sqrt(R^2 + X_C^2) )` `E_(max) = 110 sqrt2 , R = 40 Omega, v = 12 kHz, C = 100 mu F` `I_(max)= 3.88A` B. `tan phi = (-1)/(omega CR) = - 0.2` or `phi` is NEARLY zero at high frequency .C. term being negligible at high frequency ,it ACTS like resistor . For stready dc, `v = 0` and `X_C = oo` . so it acts like an open circuit for dc. |
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