In [1]:

```
# To find percentage error
# Modern Electronic Instrumentation And Measurement Techniques
# By Albert D. Helfrick, William D. Cooper
# First Edition Second Impression, 2009
# Dorling Kindersly Pvt. Ltd. India
# Example 12-1 in Page 360
import math
# Given data
R = 1 #Resistance of the wire in ohm
R_L = 10*10**3 #Load resistance in ohm
I_supply = 50*10**-3 #power supply current in A
V_out = 1 #output of the amplifier in V
#Calculations
V_L = (V_out+(I_supply*R))*R_L/(2*R+R_L)
print "The load voltage calculated = ",round(V_L,2)
percenterror = math.ceil((V_L -V_out)/V_L*100)
print "The percentage error is about ",round(percenterror),"%, which is unacceptable in most systems"
#Result
# The load voltage calculated = 1.05
# The percentage error is about 5 %, which is unacceptable in most systems
```