## PHYS345 Electricity and Electronics

**Answer for E8-1**
The voltage divider created by the Thevenin impedance -*j*20k and the
input impedance 10k brings 45% of the Thevenin emf to the amplifier input:

10/sqrt(10^{2}+20^{2}) = 0.45

To get 99%, the Thevenin reactance must be lowered to satisfy the condition:
*R*_{in}/sqrt(*R*_{in}^{2}+*X*_{C}^{2}) = 0.99

That results in *X*_{C} =0.14 *R*_{in} = 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.