Question

A circuit consists of an 92- resistor in series with a 5.9-?F capacitor, the two being connected between the terminals of an ac generator. The voltage of the generator is fixed. At what frequency is the current in the circuit one-half the value that exists when the frequency is very large? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise

Answer 1

Z = R + (-j)/{?*C} = R + (-j)/{2*?*f*C}
     = (92)+ (-j)/{2*?*f*(5.9e-6)}
     = (92)-(j)*(26989)/{f}
here ( 1/(2*pi*5.9*10^-6))=26989)
Z = {(92)² + (26989/f)²}^(1/2)  

V = I*Z    ----->    I = V/Z     ----->   I₂/I₁ = Z₁/Z₂

When "f" is very large, impedance "Z₁" approaches{Z₁ = 120}. Thus:

I₂/I₁ = (1/2)     ----->   Z₁/Z₂ = (92)/{(92)² + (26989/f)²}^(1/2) = (1/2)

     ----->   {(92)² + (26989/f)²}^(1/2) = 184
     ----->   (92)² + (26989/f)² = 33856
     ----->    (26989/f)² = 25392
     -----> 26989/f = 159.3
     ----->     f = 26989/159.3
     ----->    f = 169.4 Hz