Category: Electrical Circuits Problems

All Solved Problems in Electrical Circuits

  • AC Circuit Analysis – Sources with Different Frequencies

    AC Circuit Analysis – Sources with Different Frequencies

    In AC circuit analysis, if the circuit has sources operating at different frequencies, Superposition theorem can be used to solve the circuit. Please note that AC circuits are linear and that is why Superposition theorem is valid to solve them.

    Problem

    Determine i_x(t) where i_s(t)=4 \cos (4t) ~ A and v_s(t)=2 \sin (2t) ~ V.
    ac circuit analysis with different source frequencies

    Solution with AC Circuit Analysis

    Since sources are operating at different frequencies, i.e. 4 \frac{rad}{s} and 2 \frac{rad}{s}, we have to use the Superposition theorem. That is to say that we need to determine contribution of each source on i_x(t). Then, the final answer is to obtained by adding the individual responses in the time domain. Please note that, since the impedances depend on frequency, we need to have a different frequency-domain circuit for each frequency.

    Contribution of the current source

    To find the contribution of the current source, we need to turn off other source(s). So, we need to turn off the voltage source. This is very similar to DC circuits that we discussed before:

    Voltage sources become a short circuit when turned off.

    Turning off the voltage source for AC steady state circuit problem containing sources with different frequencies

    Frequency domain

    We first convert the circuit to the frequency domain:
    i_s(t)=4 \cos (4t) ~ A \rightarrow I_s=4\angle 0^{\circ} (\omega = 4 \frac{rad}{s})
    L=1\text{H} \rightarrow Z_L=j \omega L=j4
    C=1\text{F} \rightarrow Z_C=\frac{1}{j \omega C}=\frac{1}{j4}=-j0.25
    R=1 \Omega \rightarrow Z_R=R=1
    i_{x_1}(t) \rightarrow I_{x_1}

    Using current divider:
    I_{x_1}=\frac{Z_C}{Z_C+R}I_s
    =\frac{-j0.25}{-j0.25+1} \times 4
    =\frac{-j0.25}{1-j0.25} \times 4
    =\frac{-j0.25 \times (1+j0.25)}{(1-j0.25)\times (1+j0.25)} \times 4
    =\frac{-j0.25 + 0.0625)}{1^2+0.25^2} \times 4
    =\frac{-j0.25 + 0.0625)}{1.0625} \times 4
    =0.235-j0.941
    =0.97\angle -76^{\circ}

    Time domain

    Conversion to time-domain:
    i_{x_1}(t)=0.97 \cos (4t-76^{\circ}) ~ \text{A}
    Please note that the sinusoidal function for i_s(t) is cosine and consequently, cosine must be used in converting I_{x_1} to the time domain.

    Contribution of the voltage source

    To find the contribution of the voltage source, the current source needs to be turned off. As mentioned before:

    To turn off a current source it should be replaced by an open circuit

    Turning off the current source for AC steady state circuit problem containing sources with different frequencies

    Frequency domain

    v_s(t)=2 \sin (2t) ~ V \rightarrow V_s=2\angle 0^{\circ} (\omega = 2 \frac{rad}{s})
    L=1\text{H} \rightarrow Z_L=j \omega L=j2
    C=1\text{F} \rightarrow Z_C=\frac{1}{j \omega C}=\frac{1}{j2}=-j0.5
    R=1 \Omega \rightarrow Z_R=R=1
    i_{x_2}(t) \rightarrow I_{x_2}

    As the inductor branch is open, this is a very simple circuit with three elements in series: R, Z_C and V_s. Therefore,
    I_{x_2}=\frac{V_s}{R+Z_C}
    =\frac{2}{1-j0.5}
    =\frac{2\times (1+j0.5)}{(1-j0.5)\times (1+j0.5)}
    =\frac{2+j}{1^2+0.5^2}
    =\frac{2+j}{1.25}
    =1.6+j0.8
    =1.789\angle 26.6^{\circ}

    Time domain

    i_{x_2}(t)=1.789 \sin (2t+26.6^{\circ}) ~ \text{A}
    (why \sin?)

    Answer

    Now that we have determined both i_{x_1}(t) and i_{x_2}(t) in time domain, we can go ahead and add them up to find i_x(t). Please note that we could not add I_{x_1} and I_{x_2} because they are not phasors with the same frequency.
    i_{x}(t)=i_{x_1}(t)+i_{x_2}(t)=0.97 \cos (4t-76^{\circ}) + 1.789 \sin (2t+26.6^{\circ}) ~ \text{A}
    signal plots - AC Circuit Analysis with superposition theorem

  • Mesh (Current) Analysis Problem

    Mesh (Current) Analysis Problem

    Solve the circuit by mesh analysis and find the current I_x and the voltage across R_2.
    mesh analysis problem

    Solution

    Mesh Analysis

    There are four meshes in the circuit. So, we need to assign four mesh currents. It is better to have all the mesh currents loop in the same direction (usually clockwise) to prevent errors when writing out the equations.
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  • Find Equivalent Impedance – AC Steady State Analysis

    Determine the driving-point impedance of the network at a frequency of 2kHz:

    Determine Impedance

    Solution

    Lets first find impedance of elements one by one:

    Resistor R

    The resistor impedance is purely real and independent of frequency.

    Z_R=R=20 \Omega

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  • Superposition method – Circuit with two sources

    Superposition method – Circuit with two sources

    Find I_x using superposition rule:
    Main cuircuit to be analyzed using superposition method

    Solution

    Superposition

    The superposition theorem states that the response (voltage or current) in any branch of a linear circuit which has more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, while all other independent sources are turned off (made zero).
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  • Find Thevenin’s and Norton’s Equivalent Circuits

    Find Thevenin’s and Norton’s Equivalent Circuits

    Find Thevenin’s and Norton’s Equivalent Circuits:
    1254-1
    Suppose that R_1=5\Omega, R_2=3\Omega and I_S=2 A.

    Solution

    The circuit has both independent and dependent sources. In these cases, we need to find open circuit voltage and short circuit current to determine Norton’s (and also Thevenin’s) equivalent circuits.
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  • Solve Using Current Division Rule

    Solve Using Current Division Rule

    Find current of resistors, use the current division rule.
    Problem 1246 (1)
    Suppose that R_1=2 \Omega, R_2=4 \Omega, R_3=1 \Omega, I_S=5 A and V_S=4 V

    Solution:
    R_2 and R_3 are parallel. The current of I_S is passing through them and it is actually divided between them. The branch with lower resistance has higher current because electrons can pass through that easier than the other branch. Using the current division rule, we get
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  • Mesh Analysis – Supermesh

    Mesh Analysis – Supermesh

    Solve the circuit and find the power of sources:
    Problem 1226 - 1
    V_S=10V, I_S=4 A, R_1=2 \Omega, R_2=6 \Omega, R_3=1 \Omega, R_4=2 \Omega.

    Solution:
    There are three meshes in the circuit. So, we need to assign three mesh currents. It is better to have all the mesh currents loop in the same direction (usually clockwise) to prevent errors when writing out the equations.
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  • Solve By Source Definitions, KCL and KVL

    Solve By Source Definitions, KCL and KVL

    Find the voltage across the current source and the current passing through the voltage source.
    Problem 1213
    Assume that I_1=3A, R_1=2 \Omega, R_2=3 \Omega, R_3=2 \Omega,I_1=3A, V_1=15 V,

    Solution
    R_1 is in series with the current source; therefore, the same current passing through it as the current source:
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  • Find Voltage Using Voltage Division Rule

    Determine voltage across R_2 and R_4 using voltage division rule.
    Assume that
    V_1=20 V, R_1=10 \Omega, R_2=5 \Omega, R_3=30 \Omega and R_4=10 \Omega
    solve using voltage division rule

    Solution:
    Please note that the voltage division rule cannot be directly applied. This is to say that:
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  • Find currents using KVL

    Find currents using KVL

    Find resistor currents using KVL.
    Resistive Circuit

    Solution:

    R_1 and V_1 are parallel. So the voltage across R_1 is equal to V_1. This can be also calculated using KVL in the left hand side loop:

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