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Assignment 1(Fall 2012)  

 (Solution)                       

Circuit Theory (Phy301)

Marks: 25

Q 1:  

Find the equivalent resistance RT of given circuit. Write each step of the calculation to get maximum marks. Draw the circuit diagram of each step otherwise you will lose your marks.  

 

 

Sol.(MISSING IMAGES are in the solution file)

Step-I: 

 

Since 20Ω is parallel with 30Ω, their combined effect is

20Ω ║ 30Ω = (30 x 20)/(30+20) = 600/50 = 12Ω

The circuit adopts the form as

                             

Step-II:

 

Now 12Ω is in series with 24Ω, so their combined effect = 12Ω + 24Ω = 36Ω

Step-III:

 

Here 12Ω and 36Ω are in parallel,

so 12Ω || 36Ω = (12 x 36)/(12+36)

                        = 432/48 = 9Ω

Step-IV:

 

Since 9Ω and 18Ω are in series, so their combined effect = 9Ω+18Ω = 27Ω

Step-V:

 

Here 20Ω and 27Ω are in parallel, so

20Ω || 27Ω = (20 x 27)/(20+27) = 540/47 = 11.5Ω

 

Step-VI:

Since 12Ω and 11.5Ω are in parallel, so

12Ω || 11.5Ω = (12 x 11.5)/(12+11.5) = 138/23.5 = 5.9Ω

Here 8Ω and 5.9Ω becomes in series, so their combined effect = 8Ω+5.9Ω = 13.9Ω

Hence equivalent resistance RT of given circuit = 13.9Ω

Q 2:

Determine VR1, R2, and R3 for the given circuit. Mention the units of calculated value.

Sol.

                    Here converting current value12.3mA into Ampere A, we have

                    

Since all resistors are in series, so same value of current 0.0123A flows through each resistance

So

Current flowing through resistance R1 (82Ω) is 0.0123A, therefore voltage drop is

                     

To find R2, we first need to find voltage drop across R2 that is. Therefore we have that is

                     

Now

                                                 

Finally

                      

 Q 3:

      Answer the following questions.  

 

  1.             I.     If you wish to increase the amount of current in a resistor from 100mA to 150mA by changing the 20V source, by how many volts should you change source? To what new value should you set it?

 

  1.           II.      How much power is produced by 500mA of current through a 4.7kΩ resistor?

 

  1.         III.     There are four equal value resistors in parallel with a 12V source, and 4mA of current from the source. What is the value of each resistor?

 

Sol.

I.        Here, let

                   

            Using, we have

                       

Now if V2 represents the changed volts in the source then using 

                   

            So we’ll set the source at 30 V.

                   

II.

         

          Here

                   

          We know that

                   

 

III.

          Here

                   

          Since the resistors are in parallel, so the voltage along all the resistors shall remain same but current will be equally distributed. The following illustrates the idea

 

 

                                               R1                   R2                    R3                     R4

 

Therefore

                                

                             

          Now using, we have

                                   

            Similarly

        

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Assignment 2(Fall 2012)
(Solution)                          
Circuit Theory (Phy301)
Marks: 20

(MISSING IMAGES are in the solution file)

Q 1:   
Find the current through all meshes of given circuit. Label and identify each mesh. Draw the circuit diagram of each step otherwise you will lose your marks.   
 


Solution:
Since a current source share two meshes, so we can assign meshes and super mesh as  


Here
 
 KVL equation for mesh  
                                 
Writing KVL equation for super mesh:
 
KVL equation for  
 
Multiplying equ. 1 by 5 and equ 2 by 7 & adding both , we have
 
 
Putting in equation (2)
 
So the calculated mesh currents are
 

(MISSING IMAGES are in the solution file)

Q 2:
Determine V0 for the given circuit using nodal analysis. Label each mesh otherwise you will lose your marks.   


Sol:
There exists a voltage source between node V1 and V2, so labeling for node and super node we have circuit
 
Constraint Equation of super node is:
V1– V2 = -12V     OR
V2 – V1 = 12V………….(1)
KCL equation for Super Node is as:
(V1 – 6) / 5K + V1 / 5K + V2 / 20K= 1mA
Taking L.C.M 20 and solving we have
(4V1 – 24 + 4V1 + V2) / 20K = 1mA
8V1 + V2 = 24 + 20
8V1 + V2 = 44……………(2)
Multiply equation 1 by 8, to make it equal as equation 2, we get:
8V2 – 8V1 = 96V, and adding in equation (2) we have
    8V2 – 8V1 = 96V
    V2 + 8V1 = 44              (+)
---------------------------------------------------
    9V2 = 140V
    9V2 = 140V
    V2   = 140 / 9 V.
    V2 = 15.55V.
This is the voltage drop across 20 K node (sum of series 10k &10k)
So the voltage drop across 10K can be calculated by voltage divider rule. Hence
              V10K = (15.55x10/20
              V10K = 7.775V

Attachments:

         Assignment 3 (Fall 2012)     
                             (Solution)                          
Circuit Theory (PHY301)
Marks: 20
Due Date: Jan 16, 2013

Q.1:
Using the Source transformation method, find Vo 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: (MISSING IMAGES are in the solution file)
We want to calculate the voltage Vo using source transformation method. We proceed as
  is in series with  , so their combined effect is equal to  .

 is in series with 12 volts battery. So it can be converted into a current source of value = 12v / 3k  (by ohm’s law), which is equal to 4mA.                 

Now voltage source has transformed in the current source and the modified circuit 4mA current source and resistor 3k  will be in parallel. So modified circuit will be as,


Now we will combine current sources as 4m – 1m = 3mA, so our modified circuit will be as,

3mA source is parallel with 3k  resistor. So it can be converted into a voltage source. By using ohm’s law we have the value of voltage source = 9 volts which is in series with a resistor of 3k , so our modified circuit will be,

Positive terminal of the 9 volts battery is connected with the negative terminal of
6 volts battery so they will be summed up as shown in the circuit below,

Now 3k resistor is in parallel with 6k resistor so
3k  || 6k  = (3k  x 6k ) / 3k  +6k
          = 2k
So voltage drop across 2k? will be the same voltage as of parallel 3k and 6k ohm.
 

Now applying voltage division rule,
 

Q.2:    
Find IO in the network given below using Norton’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.


Solution: (MISSING IMAGES are in the solution file)
As we have current dependent voltage source, so first we find Vth              
 Removing load resistances to calculate Vth                     

Using mesh analysis to find Vth
 
KVL for loop 1,
 

HERE I2 = -4mA
      

Ix = I1 – I2
Ix = -2.33 mA – [-4mA]
 Ix = 1.66 mA

And according to ohm’s Law:
V2k = Ix(2 k)
V2k = (1.66 mA)(2 k)
V2k = 3.33 V    

VTH = V2k + 2kIx          
Here
Ix = 1.66 mA
VTH = 3.33 V + 2k[1.66 mA]
VTH = 3.3 V + 3.3 V
  VTH = 6.6 V   
Calculating ISC

                                                        2000Ix
by nodal analysis     V1                                                             V2

    
    6 k?        4 k?


    2 k?    4 mA                    ISC        
6 V
Ix I                                  Ix   
Equation for super node:
V1 + 6    V1         V2
    +      +      = 4 mA
4 k         2 k        4 k

3V1 + 6 + V2               
                                        = 4mA
                 4 k

3V1 + 6 + V2 = (4 mA)(4 k)

   3V1 + V2 = 10                  ………… (A)
Constraint Equation:
                                    V2 – V1 = 2000Ix
        V1
Ix =
        2 k
                            V1
V2 – V1 = 2000
                            2 k

V2 – V1 = V1
V2 – V1 - V1 = 0
V2 – 2V1 = 0
V2 = 2V1
 V1 = 0.5V2
Substituting the value of V1 in equation (A)
3[0.5V2] + V2 = 10
1.5V2 + V2 = 10
2.5V2 = 10
 V2 = 4 V
So           V2
ISC =
    4 k    

           4 V
ISC =
          4 k

ISC = 1mA
             VTH
RN =
             ISC
            6.6 V
RN =
            1mA                              RN = 6.6 k?    

Norton’s equivalent circuit is:               

        
According to current divider rule:                           
    
           6.6 k
I4k =                  × 1mA              
          6.6 k + 6 k              
    

I6k = 0.52348 mA

Attachments:

         Assignment 4 (Fall 2012)
                                   (Solution)                             
Circuit Theory (PHY301)
Marks: 25
Due Date: Feb 01, 2013

Q.1:
Calculate the peak load current for the circuit shown below.


Sol: (MISSING IMAGES and FORMULAS are in the solution file)
Here the input voltage is given an rms value. So this value is converted to a peak value                                   V1(pk) = V1(rms)/0.707
                                                = 150Vac/0.707
                                         V1(pk)  = 212.16 Vpk
        
                                         We can now find V2(pk) as
                                  V2(pk) = N2/N1  V1(pk)
                                             = (5/10)(212.16 Vpk)
                                          V2(pk) = 106Vpk

    And the load voltage is:
                                          VL(pk) = V2(pk) – VF  ( where VF is forwarding voltage)
                                                = 106 – 0.7
                                      VL(pk)= 105.38Vpk
    Now the peak load current will be
                                       IL(pk) = VL(PK)/RL
                                                     = 105.38 Vpk/5k
                                            IL(pk) = 21mApk

Q.2:
A silicon diode has n=1.Determine the change in voltage if the current changes from 0.5mA to 15mA.   
Sol: (MISSING IMAGES and FORMULAS are in the solution file)
The general diode equation in forward biased is given as

                  i = Is eV/nVT         (Where VT is constant and at room temp. its value is 25mv)
For varying currents I1 and I2
                  I1 = Is eV1/ nVT
                   I2   = Is eV2/ nVT
                   I1/I2 = e(v1 –v2)nVT
             Which can be written as
                       V1–V2 = nVTln(I1/I2)………….(A)
        Here,
                 VT= 25 mV     n=1    I1=15mA   and I2=0.5mA
             Putting in (A)
                               V1–V2 = 1 x25 ln( 15/0.5)
    So the change in the voltage will be     
                          V1–V2 = 85mV                                                                                                             
Q.3:
Give the brief answers;   
1)    Name the two possible failures in a transformer, state the most likely one.
Ans:
    (i) Open winding is a most common transformer failure; short windings are much less common.
          (ii)Operating the transformer above rated values will also cause transformer failure.

2)    Name any four semiconductor devices and their use in electronic industry.
Ans:
Semiconductor devices are made by semiconductor material and are manufactured both as single discrete devices and as integrated circuits (ICs), which consist of few (as low as two) to billions—of devices manufactured and interconnected on a single semiconductor substrate, or wafer.
Following are given some important semiconductor devices
•    Electronic transistor, junction transistor, transistor  used for amplification
•    Junction rectifier, semiconductor diode, diode used for rectification.
•    Microprocessor chip, silicon chip, chip used in computers
•    Laser diode used for optical storage, laser pointers, fiber optics, surgery.
•    Light-emitting diode (LED)  used as indicated lamps
•    Solar cell used for converting solar energy to electrical energy.
3)    Usually resistance of a conductor increases by increasing the temperature, how the conductivity of semiconductor is affected at increased temperature?
        Ans:
              By increasing temperature the conductivity of semiconductor increases due to            breaking of covalent bonds and availability of more free electrons.                                            

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