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    1) Thevinin Equivalent using source equivalent method
    2) Superposition Theorem

    © BrainMass Inc. brainmass.com December 24, 2021, 11:25 pm ad1c9bdddf
    https://brainmass.com/engineering/electronic-engineering/circuit-network-analysis-568413

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    SOLUTION This solution is FREE courtesy of BrainMass!

    As circuit diagrams are involved in the explanation please FIND ATTACHMENT to refer complete solution.

    1. SOLUTION BY THEVININ EQUIVALENT CIRCUIT

    Analysis of the linier DC circuit by Thevinin's theorem involves two main steps, (i) Finding equivalent voltage across load terminals when load resister is kept open, this equivalent voltage is called Thevinin voltage (VTH), it is also called open-circuit-voltage; and (ii) Finding equivalent resistance across load terminals when load resister is kept open and all sources are removed, this resistance is called Thevinin Resistance (RTH).

    (i) Finding equivalent voltage across load terminals when load resister is kept open, this equivalent voltage is called Thevinin voltage (VTH):

    On removing load resistance the circuit will be like below.

    Here, VTH = VCD ( As no current will flow through R3 (5 k-ohm) resister)

    {Circuit Diagram, See attachment}

    In order to calculate equivalent voltage first we should consider loop 'EGHF' on the left. Both branches GH & EF resisters of 10 k-ohms are connected to parallel to each other. So the current 2mA applied by current source will be divided by 2 for branch GH & EF giving current 1mA trough both the branches. Now in figure we observe see that the voltage applied by current source is opposite to V1, V2 and V3 as current 2mA is flowing in opposite direction through V3 & 1mA trough V1& V2, in a manner such that these tree sources are being charged by current source. And current flowing through V3 is 2mA (i.e. I = 2 mA) and through V1, V2, R1 & R3 is 1mA(i.e. i = 1 mA).

    We can now use the power conservation low, Input Power provided by current source is equals to the Output Power by combined consumption of V1, V2, V3, R1 & R2.

    That means,

    Power (Current Source) = Power(V1) + Power(V2) + Power(V3) + Power(R1) + Power(R2).

    Veq(I) = V1(i) + V2(i) + V3(I) + R1(i)2 + R2(i)2 {Where, Veq - Voltage Across Current source = VCD and i = 0.001 A and I = 0.002 A}
    Veq(0.002) = 10 (0.001) + 20 (0.001) + 5 (0.002) + 10000 (0.001)2 + 10000 (0.001)2
    {Where, V1=10V, V2=20V,V3=5V,R1=R2=10k-ohm=10000 ohm, i = 1 mA =0.001 A & I = 2mA=0.002 A }
    Veq(0.002) = 0.01+0.02+0.01+0.01+0.01=0.06
    Veq=0.06/0.002=30 V

    So VTH = VCD=Veq=30 V ( as VTH = VCD)

    (ii) Finding equivalent resistance across load terminals when load resister is kept open and all sources are removed, this resistance is called Thevinin Resistance (RTH)

    Here, removing of voltage source means replacing it by short as shown in figure and removing current source means keeping its terminals as open.

    {Circuit Diagram, See attachment}

    It is easy to observe that both 10 k-ohm resisters are parallel to each other and 5 k-ohm is in series with this equivalent resister.

    So we will get,

    RTH = 5 k-omh + [(10*10)/(10+10)] k-ohm = 5 + 100/20 = 5 + 5 = 10 k-ohm.

    Therefore, RTH = 10 k-ohm

    So the final Thevinin's equivalent circuit will have voltage source of 30 V and resistance of 10 k-ohm in series with load resister as below,

    {Circuit Diagram, See attachment}

    Now it current through above circuit, I = 30/60 = 0.5 mA = 0.5 x 10-3 A. So load across RL will be
    VL = 50 x 103 * (0.5 x 10-3) = 25 V

    Hence, VL = 25 V

    2. SOLUTION BY THE SUPERPOSITION THEOREM

    When applying superimposition theorem, we will find load voltage for only one source by removing all other sources and will repeat this process for all the sources. Finally we will combine results from all four sources V1, V2, V3 and Current Source in order to find actual VL for the given circuit.

    FOR V1 : By removing V2, V3 and Current source.
    Removing of voltage source implies replacing it with short and removing current source implies replacing it with open terminals as below.

    {Circuit Diagram, See attachment}

    In above circuit 5 k-ohm and 55 k-ohm are in parallel to 10 k-ohm and one 10 k-ohm resister (left) is in series with this parallel equivalent resistance, Giving equivalent resistance in circuit as,
    Req = 10 + [(10*55)/65] = 10 + (550/65) = 10 + (110/13) = 240/13 k-ohm = (240/13) x 103. So total current I in the circuit will be, I = 10/[(240/13) x 103] = [(10 *13)/ 240] x 10-3 = 130/240 x 10-3 = 13/24 mA.
    In order to find VL,1 we will need to find current passing through load resister.
    Now applying Kirchhoff's rule to loop 'EABF',

    10 x 103 (I1) = 55 x 103 (I2), (but by junction rule I = I1+I2 i.e. I1 = I - I2.)

    Therefore,
    10 x 103 ( I - I2 ) = 55 x 103 ( I2)
    Or, 10 x I = 65 x I2
    Or I2 = (10 * 13)/(24*65) = (2 * 5 * 13)/(24*65) = 1/12 mA = (1/12) x 10-3 A.

    Therefore, VL,1 = 50 x 103 x I2 = 50/12 V.

    FOR V2 : By removing V1, V3 and Current source.
    On keeping V2 as it is and removing V1, V3 and CurrentSource; Simillar to the calculations for above case of V1 = 10V the equivalent resistance will remain same. The only difference is the source of twise voltage 20 V.
    So I2 will be 1/6 mA.

    Therefore, VL,2 = 50 x 103 x I2 = 50/6 V.

    FOR V3 : By removing V1, V2 and Current source.

    {Circuit Diagram, See attachment}

    Here, I1 = I2 as both the resisters have same value. As both 10 k-ohms are parallel to each other equivalent resistance of them will be = 10*10/(10+10) = 5 k-ohm. This equivalent resister is in series with 5 & 55 k-ohms of right side. Giving final equivalent resistance of circuit = 5 + 50 +5 = 60 k-ohms.

    Therefore, current, I = 5/60 mA = 1/12 mA.
    'I' is the current passing trough load resister,

    Hence,
    Load voltage, VL,3 = 50 x 103 x I = 50/12 V.

    FOR CURRENT SOURCE : By removing V1, V2 and V3.

    I1 & I2 current is passing trough left and right loops respectively. Here, I = I1 + I2 = 2 mA = 0.002 A. Using Kirchhoff's rule for left & right loop of the current source, We get,

    5 k-ohm x I1 = Vs & 55 k-ohm x I2 = Vs

    Therefore,
    5 k-ohm x I1 = 55 k-ohm x I2
    Or 5 x 103 x (I-I2) = 55 x 103 x I2
    Or 5 x 103 x (0.002-I2) = 55 x 103 x I2
    Or 10 - (5000xI2) = 55000 x I2
    Or 60000 x I2 = 10
    Or I2 = 10/60000 = (1/6) x 10-3 = 1/6 mA

    Now, as current I2 = 1/6 mA is passing through load resister RL, Load voltage, VL,4 will be,

    VL,4 = 50 x 103 x I2 = 50/6 V

    Combining results to find actual VL:
    Actual VL will be superposition of all the results obtained for the individual sources.
    VL = VL,1 + VL,2 + VL,3 + VL,4
    = 50/12 + 50/6 + 50/12 + 50/6
    = (50 + 100 +50 +100)/12
    = 300/12
    VL = 25 V

    Hence you can conclude that the voltage across load is 25 V and we get same result by either both approaches.

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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 11:25 pm ad1c9bdddf>
    https://brainmass.com/engineering/electronic-engineering/circuit-network-analysis-568413

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