Fall 2011_PHY301_1_SOL

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					Assignment 1(Fall 2011)
(Solution)
                                                                   Circuit Theory (Phy301)
                                                                                 Marks: 25
                                                                            th
                                                               Due Date: 27 October, 2011


Q.1
Find the equivalent resistance at point a, b 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.




Solution:
       Step 1:
                 2Ω resistor is in series with 2Ω resistor
                 So 2Ω + 2Ω = 4Ω




       Step 2:
                 4Ω resistor is in parallel with 4Ω resistor
                 So      1/R = 1/R + 1/R
                             eq        1       2
                                  = 1/4Ω + 1/4Ω
                                  = (4 × 4)/(4 + 4)
                                  = 16/8
                                  = 2Ω
Step 3:
          6Ω Resistor is in series with 2Ω Resistor
          So 6Ω + 2Ω = 8Ω




Step 4:
          8Ω resistor is in parallel with 8Ω resistor
          So       1/Req = 1/R1 + 1/R2
                         = 1/8Ω + 1/8Ω
                         = (8 × 8)/(8 + 8)
                         = 64/16
                         = 4Ω




Step 5:
          5Ω resistor is in series with 4Ω resistor
          So 5Ω + 4Ω = 9Ω
Q.2
Using the division rule, find V and I mentioned in the given circuit.




Sol:

Finding Current I:
        As 10Ω resistor is in series with current source therefore current remains same
        Current will divide at point Node. So current I through 4Ω can be found using
current divider formula.
                           R2              2
                     I          *12 A        *12
                         R2  R3         24
                       2      24
                       *12 
                       6      6
                    I  4A
       So current flow through 4Ω Resistor will be 4A.
       Now current flow through 2Ω resistor is
                            R3             4
                    I2           *12 A      *12
                         R3  R 2         42
                        4     48
                       *12      8A
                        6     6
                   I 2  8A
       Current flow through 2Ω resistor is 8A.

Finding Voltage V:
              As we know that
              V = IR
              So we can find voltage across 2Ω resistor as
                   V = I2 R
                     =8×2
                     = 16 V
              Voltage across 2Ω resistor is 16V.
Q.3
For the given circuit below, find current flow through each element, for the case when
   (a) Point a, b is open.
   (b) Point a, b is closed.




   Solution:
      (a) When Point a, b is open
          As we know that P = V × I
          So       I = P/V
                     = 4/20
                     = 0.2A
          So current I1 is 0.2A
             As point a, b is open therefore current I2 and I3 will be zero
       (b) When point a, b is closed
       First we see that all elements are in parallel, so same 20v voltage drops across
       each element.
           As we know that P = V × I
           So        I = P/V
                       = 4/20
                       = 0.2A
           So current I1 is 0.2A
Finding I2
                As    V = IR
                       I = V/R
                      I2 = 20/5000
                      I2 = 0.004A
                So current I2 is 0.004A=4mA
Finding I3
                As P=VI
                    I = P/V
                   I3 = 10/20
                      I3 = 0.5A
                So current I3 is 0.5A

				
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Description: Past assignments, CS201, CS301, CS304, CS601, Pak301, Eng201, PHY301