We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
+ Link For Assignments, GDBs & Online Quizzes Solution |
+ Link For Past Papers, Solved MCQs, Short Notes & More |
Dear Students! Share your Assignments / GDBs / Quizzes files as you receive in your LMS, So it can be discussed/solved timely. Add Discussion
How to Add New Discussion in Study Group ? Step By Step Guide Click Here.
Circuit Theory (PHY301)
Marks: 30
Due Date: January 6, 2011
DON’T MISS THESE Important instructions:
· To solve this assignment, you should have good command over first 27 lectures.
· Upload assignments properly through LMS, (No Assignment will be accepted through email).
· Write your ID on the top of your solution file.
· All students are directed to use the font and style of text as is used in this document.
· Don’t use colorful back grounds in your solution files.
· Use Math Type or Equation Editor etc for mathematical symbols.
· This is not a group assignment, it is an individual assignment so be careful and avoid copying others’ work. If some assignment is found to be copy of some other, both will be awarded zero marks. It also suggests you to keep your assignment safe from others. No excuse will be accepted by anyone if found to be copying or letting others copy.
· Don’t wait for the last date to submit your assignment.
· You can draw circuit diagrams in “Paint” or in “Corel Draw”. The simple and easy way is to copy the given image in Paint and do the required changes in it.
Q.1.
Using the superposition method find V_{o} in the network given below. Label and draw each step where it required. Write each step of the calculation to get maximum marks and also mention the units of each derived value.
Q.2
Using the source transformation method find V_{o} in the network given below. Label and draw each step where it required. Write each step of the calculation to get maximum marks and also mention the units of each derived value.
Q.3
Find V_{O} in the network given below using Thévenin’s theorem.
Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.
Tags:
+ How to Follow the New Added Discussions at Your Mail Address?
+ How to Join Subject Study Groups & Get Helping Material? + How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators? + VU Students Reserves The Right to Delete Your Profile, If?.
+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)
+ Click Here to Search (Looking For something at vustudents.ning.com?) + Click Here To Join (Our facebook study Group)Marks: 30
Due Date: January 6, 2011
.
Q.1.
Using superposition to find a given output can be broken down into foursteps:
1. Isolate a source -Select a source, and set all of the remaining sources to zero. The consequencesof "turning off" these sources are explained in Open and ClosedCircuits. In summary, turning off a voltage source results in ashort circuit, and turning off a current source results in an open circuit.(Reasoning - no current can flow through a open circuit and there can be novoltage drop across a short circuit.)
2. Find the output from the isolated source - Once asource has been isolated, the response from the source in question can be foundusing any of the techniques we've learnedthus far.
3. Repeat steps 1 and 2for each source - Continue to choose a source,set the remaining sources to zero, and find the response. Repeat this procedureuntil every source has been accounted for.
4. Sum the Outputs -Once the output due to each source has been found, add themtogether to find the total response.
Solution :
V =V +V .........................(1)
Applying KCL at node V1
V V −V V 1 + 1 2 + 1 =0 4k 2k 12k 3V + 6V − 6V + V1 1 2 1 =0 12k 10V − 6V = 0 ...........................( 3) S
V =V2 -V1
In the circuits,
..............................(2)
Consider only 10Vvoltage source, We ignore effect of 4mA and open circuit it to find V01.V01 can be determined by voltage divider, loop method ornodal analysis, we use nodal analysis here
Solution: As we knowin superposition method weconsider the effect of each source one by one and ignore othersource and then add result of each source to find v0By using superposition principle to the given circuit
Applying KCL at node V2
4V − 2V = 2
.................( 4 )
Solving equations (3) & (4) we can get,
V = 2.143V & 1 V = 3.5V 2
From ………… (2)
In the above circuit,
Applying KCL at node V1
V V −V V 1 + 1 2 + 1 + (4 ×10−3) = 0 4k 2k 12k 3V+ 6V − 6V + V + 48 1 1 2 1 =0 12k 10V − 6V = −48............................(6)
V = V2 −V1.................(5)
Consider only 4mAcurrent source, i.e., 10V is short circuited and find V02 value. V02 can be determined by currentdivider, loop method or nodal analysis, we use nodalanalysis here
V = 3.5 − 2.14 01 ∴V = 1.4V 01
V −V V −10 V 2 1+ 2 + 2 =0 2k 4k 4k 2V − 2V + V−10 + V 2 1 2 2 =0
Applying KCL at node V2
V= 4.8V approxi.
Q.2
Solution :
by source transformation
Now2mA source is in parallel with 3k resistor.So itcan be changed to a voltage source of value = 2m x 3k (by ohm’s Law)
=6 Volts.
3kresistor will become in series withthis source as shown in the circuit below
Nowthe combined effect of these two source will be 9 volts.
So
Vth= 9 voltsThird step:
Calculating Rth Rth =3kFourth step:
Calculatingthe unknown quantity.
Aftercalculating Vth and Rth, re-inserting the load resistance RL in thecircuit in series with Rth andconsidering the Vth as a battery in series withthese two resistances.
V0 = (6k/9k) x9 V0= 6voltsQ.3
Solution :
Removing load resistance RL= 1KΩ
In the above circuit Ix= 0 because current is not completing its path due to open circuit so, value ofdependent current source is also zero.
VTH = 6 V Now for RTHwe will shortthe open terminal of the Thévenin’s circuit andcalculate the Isc and then divide VTH with Isc to calculate RTH
Applying KCL at node VSum of all the currents leaving the junction = sum of all the currents enteringthe junction V V+6 + = 2Ix 1k 2k Here V+6 Ix = 2k V + 1k 2k V+6 =2 2k V+6
V + 1k V + 1k
V+6 = 2k V+6 2k
V+6 1k V+6 =0 1k =0
2V + V + 6 – 2[V + 6] 2k 2V + V + 6 – 2[V + 6] = 02V + V + 6 – 2V - 12 = 0 V=6V Now 6+6 Ix = 2k
Ix = 6 mA = ISC RTH = VTH ISC 6V RTH = 6 mA
RTH = 1 kΩ
THEVENIN’S EQUIVALENT:
According to Voltage divider rule: 1k V0 = ×6V1k+1k
V0 = 3 V
© 2020 Created by + M.Tariq Malik. Powered by
Promote Us | Report an Issue | Privacy Policy | Terms of Service
VU Students reserves the right to delete profile, which does not show any Activity at site nor has not activity more than 01 month.
We are user-generated contents site. All product, videos, pictures & others contents on vustudents.ning.com don't seem to be beneath our Copyrights & belong to their respected owners & freely available on public domains. We believe in Our Policy & do according to them. If Any content is offensive in your Copyrights then please email at m.tariqmalik@gmail.com or Contact us at contact Page with copyright detail & We will happy to remove it immediately.
Management: Admins ::: Moderators
Awards Badges List | Moderators Group
All Members | Featured Members | Top Reputation Members | Angels Members | Intellectual Members | Criteria for Selection
Become a Team Member | Safety Guidelines for New | Site FAQ & Rules | Safety Matters | Online Safety | Rules For Blog Post