VU VICKY Virtual University Assignment Solution, Past Papers, Handouts and other material

  

Assignment 3(Spring 2018)

                                Complete sol is in file below            

Circuit Theory (Phy301)

Marks: 25

Due Date: July 23, 2018

DON’T MISS THESE Important instructions:

         

·          To solve this assignment, you should have good command over first 28 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.  

  

·          You can draw circuit diagrams in “Paint” in “Corel Draw” or in “circuit maker”. 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 voltage drop V_o across 3KΩ 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.        

                     

Solution: Applying principle of superposition, we will take effects of the sources one by one:

Current Source:

2kΩ and 2kΩ are both parallel so 2kΩ||2kΩ=1kΩ ,

So the circuit becomes:

Using current division rule, current through 3kΩ will be:

I₀₁ = (6×1)/ (1+3+1)= 6/4 = 1.5 mA

Voltage Source:

                                       

3k and 1k are in series so:

3k+1k=4k

So the new circuit is: 

                                         

4k and 2k are in parallel, so;

4k||2k=1.33k

So the new circuit is:

                                                            

Applying voltage division rule:

V_1.33k = (4×1.33)/ (2+1.33) = 5.32/3.33=1.598kV

 As 1.33k is the resultant of 2k and 4k, so same voltage will drop there:

So I₀₂ = V/1.33 = 1.598/1.33 = 1.201 mA

Same current flows through 3k and 1k in series, so

I₀ = I₀₁+I₀₂=1.5+1.201= 2.7 mA

Now V₀ through 3k resistor will be:

V₀ = I₀R = 2.7 mA × 3kΩ = 8.1 V

V₀₌8.1 V Answer

                                                                                   

Q: 2:

Find R_th (Thevenin’s resistance ), V_th (Thevenin’s Voltage)  and I_N (Norton’s current) across a, b. Draw equivalent circuit and find Vo across 5Ω. Draw and label the circuit diagram of each step and also mention the units of each derived value.

    Ans.

Find Thevenin's resistance

<<Download>>