# Chapter 1 - ELECTRONIC MATERIALS AND COMPONENTS¶

## Example E1 - Pg 43¶

In [1]:
#Calculate the value of resistor
print '%s' %("Refer to the figure 1.52")
print '%s' %("Hold the resistor as shown in the figure such that tolerance is on your extreme right.")
print '%s' %("Now the value of the resistor is equal to")
print '%s' %("  Red      Black       Blue        Gold")
print '%s' %("   2         0          6         (+/-)5%")
red=2. #red value
blk=0 #black value
blu=6. #blue value
gld=5. #gold value
value_res=(red*10.+blk)*10.**blu #value of resistor
print '%s %.0f %s %.0f' %("\n The value of resistor is",value_res,"ohm (+/-)",gld)
per_val=0.05*value_res
pos_value_res=value_res+per_val #positive range of resistor
neg_value_res=value_res-per_val #negative range of resistor
print '%s %.0f %s %.0f %s ' %("\n The value of resistor is",neg_value_res*1e-6," Mohm and",pos_value_res*1e-6,"Mohm")

Refer to the figure 1.52
Hold the resistor as shown in the figure such that tolerance is on your extreme right.
Now the value of the resistor is equal to
Red      Black       Blue        Gold
2         0          6         (+/-)5%

The value of resistor is 20000000 ohm (+/-) 5

The value of resistor is 19  Mohm and 21 Mohm


## Example E2 - Pg 43¶

In [2]:
#Calculate the value of the resistor
print '%s' %("With the help of colour coding table, one finds")
print '%s' %("  1st_Band    2nd_Band    3rd_Band    4th_Band")
print("    Yellow      Violet      Orange      Gold")
print '%s' %("      4          7         10**3     (+/-)5%")
yel=4. #yellow value
vio=7. #violet value
org=1e3 #orange value
gld=5. #gold value in %
val_res=(yel*10.+vio)*org
print '%s %.2f %s %.0f' %("\n The value of resistor is",val_res*1e-3,"kohm (+/-)",gld)
gld_ab=0.05 #absolute gold value
per_val=gld_ab*val_res
print '%s %.0f %s'%("\n Now, 5%% of 47k_ohm =",per_val, "ohm")
range_high=val_res+per_val #higher range
range_low=val_res-per_val #lower range
print'%s %.2f %s %.2f %s %.2f %s %.2f %s'%("\n\n Thus resistance should be within the range",val_res*1e-3, "kohm(+/-)",per_val*1e-3 ,"Kohm\n or between",range_low*1e-3 ,"kohm and",range_high*1e-3, "kohm")

With the help of colour coding table, one finds
1st_Band    2nd_Band    3rd_Band    4th_Band
Yellow      Violet      Orange      Gold
4          7         10**3     (+/-)5%

The value of resistor is 47.00 kohm (+/-) 5

Now, 5%% of 47k_ohm = 2350 ohm

Thus resistance should be within the range 47.00 kohm(+/-) 2.35 Kohm
or between 44.65 kohm and 49.35 kohm


## Example E3 - Pg 44¶

In [3]:
#Calculate the value of the resistor
print '%s' %("The specification of the resistor from the color coding table is as follows")
print '%s' %("  1st_Band    2nd_Band    3rd_Band    4th_Band")
print '%s' %("     Gray        Blue        Gold       Silver")
print '%s' %("      8          6        10**(-1)     (+/-)10%")
gray=8. #gray value
blu=6. #blue value
gld=10.**-1 #gold value
sil=10. #silver value in %
val_res=(gray*10.+blu)*gld
print '%s %.2f %s %.2f' %("\n The value of resistor is",val_res, "ohm (+/-)",sil)
sil_ab=0.1 #absolute gold value
per_val=sil_ab*val_res
print '%s %.2f %s' %("\n Now, 10%% of 8.6 ohm =",per_val," ohm")
range_high=val_res+per_val #higher range
range_low=val_res-per_val #lower range
print '%s %.2f %s %.2f %s %.2f %s %.2f %s' %("\n\n Obviously resistance should lie within the range",val_res,"ohm(+/-)",per_val,"ohm\n or between",range_high,"ohm and",range_low,"ohm")

The specification of the resistor from the color coding table is as follows
1st_Band    2nd_Band    3rd_Band    4th_Band
Gray        Blue        Gold       Silver
8          6        10**(-1)     (+/-)10%

The value of resistor is 8.60 ohm (+/-) 10.00

Now, 10%% of 8.6 ohm = 0.86  ohm

Obviously resistance should lie within the range 8.60 ohm(+/-) 0.86 ohm
or between 9.46 ohm and 7.74 ohm


## Example E4 - Pg 44¶

In [4]:
#Calculate the load voltage and load current
print '%s' %("Refer to the figure 1.53")
Vs=2. #supply voltage in V
Rs=1. #resistance in ohm
Is=Vs/Rs
print '%s %.2f %s' %("\n Current Is =",Is," A \n")
print '%s' %("     Internal resistance remains the same but is now connected in parralel with the current sourceIS,as shown in Figure 1.51(a)")
print '%s' %("    Now,we connect a load resistance R_L=1 ohm across the terminals of two representations ,and find I_L and V_L.  From Figure 1.54(b) and using the current-divider concept,one obtains")
print '%s %.0f %s' %("\n Load voltage =",VL,"V")
print '%s %.0f %s' %("\n Load current =",IL,"A \n")
print '%s' %("From equation 53(b),using the voltage-divider concept,one obtains")
VD_vl=Vs*(RL/(RL+Rs)) #load voltage using voltage divider
print '%s %.0f %s' %("\n Load voltage =",VD_vl,"V")
print '%s %.0f %s' %("\n Load current =",VD_il,"A \n")

Refer to the figure 1.53

Current Is = 2.00  A

Internal resistance remains the same but is now connected in parralel with the current sourceIS,as shown in Figure 1.51(a)
Now,we connect a load resistance R_L=1 ohm across the terminals of two representations ,and find I_L and V_L.  From Figure 1.54(b) and using the current-divider concept,one obtains

From equation 53(b),using the voltage-divider concept,one obtains



## Example E5 - Pg 45¶

In [5]:
#Calculate the percentage variation in current, current for two extreme values R_L
print '%s' %("Refer to the figure 1.55")
print '%s' %("(a) R_L varies from 1 ohm to 10 ohm.")
print '%s' %("Currents for two extreme values of R_L are")
Vs=10. #supply voltage
RL1=1. #resistance RL1
Rs=100. #source resistance
IL1=(Vs/(RL1+Rs))
RL2=10.
IL2=(Vs/(RL2+Rs))
per_var_cur=((IL1-IL2)/IL1)*100.
print '%s %.2f %s' %("\n Percentage variation in current =",per_var_cur,"\n")#answer in the text book took a .3 decimal round off value
print '%s' %("   Now,load voltage for the two extreme values of R_L are")
VL1=IL1*RL1
VL2=IL2*RL2
per_var_vol=((VL2-VL1)/VL2)*100.
print '%s %.2f %s' %("\n Percentage variation in current =",per_var_vol,"\n")

print '%s' %("(b) R_L varies from 1 k-ohm to 10 k-ohm(Figure 1.55(b))")
print '%s' %("Currents for the two extreme values R_L are")
RL11=1000.
IL11=(Vs/(RL11+Rs))
RL22=10000.
IL22=(Vs/(RL22+Rs)) #mistake in book value
per_var_cur11=((IL11-IL22)/IL11)*100.
print '%s %.2f %s' %("\n Percentage variation in current =",per_var_cur11,"\n") #mistake in book value

Refer to the figure 1.55
(a) R_L varies from 1 ohm to 10 ohm.
Currents for two extreme values of R_L are

Percentage variation in current = 8.18

Now,load voltage for the two extreme values of R_L are

Percentage variation in current = 89.11

(b) R_L varies from 1 k-ohm to 10 k-ohm(Figure 1.55(b))
Currents for the two extreme values R_L are

Percentage variation in current = 89.11