Chapter 7:Parallel Circuits

Problem no:7.5, Page no:67

In [1]:
from __future__ import division

#Cal of Equivalent resistance

#Initialization

#Resistors connected in parallel in Ohm

R1=5

R2=10

from decimal import *

#Calculation

R=(R1*R2)/(R1+R2)

print "R=",round(R,2),"ohm"

R= 3.33 ohm

Problem no:7.1, Page no:64

In [8]:
from __future__ import division

#Cal of Equivalent resistance

#Initialization

#Resistors connected in parallel

R1=R2=R3=5                  #in Ohm

#Calculation

R=1/((1/R1)+(1/R2)+(1/R3))

print "R=",round(R,2),"Ohm"

R= 1.67 Ohm

Problem no:7.6, Page no:67

In [10]:
from __future__ import division

#Cal of resistance(R2) to be reduced

#Initialization

#Resistance in Ohm

R1=50

R=20

R2=(R1*R)/(R1-R)

print "R2=",round(R2,1),"Ohm"
 

R2= 33.3

Problem no:7.7, Page no:67

In [19]:
from __future__ import division

#Cal of Power in series and parallel

#Initialization

R1=R2=240                            # in Ohm

V=120                                #voltage in volts

#(a)Power in series

R=R1+R2                              #Equivalent resistance

I=V/R                                #Current in the circuit

P=(I**2)*R1                          #Power dissipated

print "(a)P=",int(P),"W"

#(b)POwer in Parallel

P=(V**2)/R1

print "(b)P=",int(P),"W"


print "The bulbs will be higher when connected in parallel."

(a)P= 15 W
(b)P= 60 W
The bulbs will be higher when connected in parallel.

Problem no:7.10, Page no:69

In [23]:
from __future__ import division

#Cal of resistor to be connected with the meter

#Initialization

I_meter=0.001        #meter current in ampere

I_shunt=0.999        #Shunt current in ampere

R_meter=40            #Shunt resistance in Ohm

R_shunt=(I_meter/I_shunt)*R_meter

print "R_shunt=",round(R_shunt,2),"Ohm"

print "A",R_shunt,"Ohm resistor in parallel with the meter will permit it to measure currents from 0 to 1 A"

R_shunt= 0.04 Ohm
A 0.04004004004 Ohm resistor in parallel with the meter will permit it to measure currents from 0 to 1 A