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PHY301 - Circuit Theory

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  • Phy301 Assignment 1 Solution and Discussion

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    @zareen said in PHY301 Assignment 2 Solution and Discussion:

    Label and identify meshes in given below circuit. Use Loop/Mesh analysis to find currents through all Meshes.

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    Spring 2020_PHY301_1_SOL.pdf

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    @zareen said in PHY301 GDB 1 Solution and Discussion:

    what types of charges are responsible for flow of current in each matter?

    Insulators and conductors can be solid, liquid or gas, and in some exceptions like glass (solid) which is an insulator becomes conductors when melted at the higher temperature. On the other hand, semiconductors are present in the solid form.

    Liquids can be conductors or insulators, depends on other properties. Though absolute pure water is an insulator, the liquid metals are electrically conductive. Gases also become electrically conductive when ionized, though they usually are insulators.

    Conductivity is the phenomenon of transmitting something like heat, electricity or sound. So, based on the conductivity of any material and the presence of a forbidden gap, they (materials) can be classified as conductors, semiconductors or insulators. In the article, we will be differentiating the three terms concerning other points on which they vary.

  • solution send kar dein

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    Q. 1 Solution:
    Starting from lower side of circuit network, we can see that 2Ω and 2Ω are in series, so their combined effect is
    2+2=4Ω
    97b5d636-b0d8-40d1-8f33-9266b3779585-image.png

    This 4Ω is in parallel of 4Ω to lower side, so their equivalent is

    1b08526d-0c95-43b2-9f73-3927652469d8-image.png

    0888b8c5-aeba-4acd-9184-9f1a8f70b1ef-image.png

    This 2Ω is in series of 5Ω and 1Ω, so the sum of series resistances is

    2Ω+ 5Ω+1Ω=8Ω
    f1ffb19b-69a3-4eb0-8d8d-4eb1316289f4-image.png

    Now at upper side, 4Ω and 2Ω are in series
    4Ω+ 2Ω=6Ω
    b3a9072f-b02e-4881-aaf3-9b389aab13e0-image.png

    This 6Ω is in parallel of 6Ω, so their equivalent is
    df9ccb62-3a78-4fec-a9d8-9044a372be75-image.png

    cac80a33-5165-4e48-b124-810aaad24d2e-image.png

    This 3Ω is in series of 3Ω and 1Ω, so the sum of series is

    3Ω+ 3Ω+1Ω=7Ω
    051e272e-1e7b-48b8-9359-4810dcb187d5-image.png

    Now we can see 7Ω becomes in series of 8Ω, so equivalent resistance is
    Req =7Ω + 8Ω
    Req=15Ω

    Q.2 Solution:
    1)
    To calculate the source current Is, Firstly, we calculate the total resistance
    RT= R1+R2+R3
    =6Ω
    V=IR
    IS=VS/RT
    = 12/6
    =2A

    2)
    Since all resistances are in series, same 2A current pass through each resistance.
    V1= R1IS
    =12
    =2V
    V2=R2Is
    =22
    =4V
    V3=R3IS
    = 32
    =6V

    3)
    P1 =I2R1
    = (2)2 *1
    =4W
    P2 =I2R2
    = (2)2 *2
    = 8W
    P3 =I2R3
    =(2)2 *3
    = 12W
    = 12W

    4)-
    Ps = VsIs
    P =122
    =24 W
    PT=P1+P2+P3 or PT=I2RT
    =4+8+12
    =24 W