PHYS345 Electricity and Electronics

Answer for E8-1

The voltage divider created by the Thevenin impedance -j20k and the input impedance 10k brings 45% of the Thevenin emf to the amplifier input:
10/sqrt(102+202) = 0.45

To get 99%, the Thevenin reactance must be lowered to satisfy the condition: Rin/sqrt(Rin2+XC2) = 0.99
That results in XC =0.14 Rin = 1.4k.

To get a reactance of 1.4k from a 500 pF capacitor requires a frequency of 227 kHz.


"http://www.physics.udel.edu/~watson/phys345/protected/exercises/answers/E8-1.html"
Last updated Oct. 30, 1998.
Copyright George Watson, Univ. of Delaware, 1998.