## PHYS345 Electricity and Electronics

**Answer for P8-17**
*A*_{v} = *V*_{2}/*V*_{t} =(10 V)/(42 mV) = 238 = 47.5dB.

Gain of 238 on an input offset of 5 mV yields an output offset of 1.2 V.

(42+5)/45 = 1.12, so the input offset makes the thermocouple emf look 12% higher than it actually is,
thus the perceived temperature is 12% higher than 1000 deg; that is, 120 deg of error.

Changes in gain amplify the thermocouple emf and the input offset in the same proportion,
so no minimization of input offset is accomplished by reducing the gain.

**Answer for P8-19**

The Thevenin resistance of (2+6.4+6.4) ohm =14.8 ohm and the input impedance of
50 ohm creates a voltage divider that reduces the amplifier input to
*V*_{1} = 50/(14.8+50) *V*_{t} = (50/64.8) 42 mV = 32 mV.

**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.

**Answer for E8-1**

The result of rationalizing Eq.1 is:

**G**_{v} = G_{0}(1-*j*x)/(1+x^{2})
where x = f/f_{u} has been introduced as a dimensionless frequency.

This expression is now in standard form and the phase angle may be extracted from the
imaginary and real parts:

tan(phase angle) = *Im*{**G**_{v}}/*Re*{**G**_{v}} = -x.

Find the phase angle for x = 0.5, 1.5, 2.5, and infinity.

"http://www.physics.udel.edu/~watson/phys345/protected/exercises/answers/1020.html"

Last updated Oct. 30, 1998.

Copyright George Watson, Univ. of Delaware, 1998.