Objective Questions on Network Analysis 11

  1. Find i(t) when VS (t) = 100 sin(2510 t).

    Zeq = 10 ω i(t) = VS / 10 = 10 sin(2510t).

  2. What is h(t) in following circuit ?

    Step response is given as [1 – e -α t]u(t)
    h(t) is an impulse response.
    \therefore h(t)=\frac{d}{dt}[1-e^{-\alpha t}]

  3. In a pure parallel LC circuit under resonance condition, current drawn from main supply is

    Current in LC circuit is 0 at resonance, because of infinite of parallel LC circuit.

  4. Z may comprise of following circuit

    Series combination of 2.5 mH & 16&mm; F will result in a lossless resonant L-C circuit with constant V.
    But parallel combination of 2.5 mH & 16 μ F will result in an exponentially increasing sinusoidal response as shown.

  5. In a RLC circuit (series) if C is increasing & remaining R & L are constants, then resonant frequency will be

    As resonant frequency, f0,
    f_0 = \frac{1}{Z\pi\sqrt{LC}}

  6. If all the transmission zeros of a network are at infinity then it is a

    lim T(s) = 0
    s → ∞
    So it will be a low pass filter.

  7. At lower half power frequency current in series R-L-C circuit has the phase relation with reference to supply voltage as

    \theta =tan^{-1}\left(\frac{X_L-X_C}{R}\right)=\tan^{-1}\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)

  8. A circuit with a resistor, inductor and capacitor in series is resonant of f0 Hz. If all the component values are now doubled the new resonant frequency is

    Resonant frequency, f0,
    f_0 = \frac {1} {2 \ pi \ sqrt {LC}}

  9. In a series RLC high Q circuit, the current peaks at a frequency

    Q = \omega_n\times \frac {L}{C} = \frac {L}{C} \frac {1} {R}
    Now high Q means low R, C & high L.
    So bandwidth = f0/Q
    Since less bandwidth, hence sharp resonance.

  10. A sinusoidal voltage V & frequency f connected to a series circuit at variable resistance R and a fixed reactance X. The locus of the tip of the current phase or I as R is varied from zero to infinity is

    I=\frac{V}{R+jX}=\frac{V(R-jX)}{R^2+X^2}
    So\;real\;axis\;current\;I_x = \frac{VR} {R^2 + X^2}
    Imaginary\; axis\; current,\; I_y = \frac{-VX} {R^2 + X^2}
    I_x^2 + I_y^2 = \frac{V^2}{R^2 + X^2} = -\frac {VI_y}{X}
    or, I_x^2\left(I_y+\frac{V}{2X}\right)^2= \frac{V^2}{4X^2}
    Above equation is a semicircle equation with radius V/2X.

  11. In the network shown below, switch S was closed & steady state is a attained. If S is opened at t = 0 & i(t) = 2 cos 100t then what is the value of V ?

    Circuit starts oscillating when switch is opened.
    Here I = 2, R = 2.5 Ω.

  12. Circuit shown below has been in steady state when switch S is opened, the current I is given by

    When the switch is closed and circuit reached its steady state condition. Since inductor offers short circuit path. So
    I = 10 / 1 = 10A
    Now, when switch is opened,
    I(t) = I_{final} + [I_{initial} – I_{final}] \ times e^ { – \frac {Rt} {L}}= 10 e^{– 2t}

  13. The current i(t) at any instant t in the network shown in the figure will be after closing the switch S at t = 0

    The value of the current at t = 0 + and at t = infinity; which should be zero and 10 / 1 = 10 A respectively.

  14. A series RLC circuit consisting of R = 20 ohms XL = 20 ohms is connected across an ac supply of 200 V rms. The rms voltage across the capacitor is

    Z=\sqrt {R^2 + (X_L–X_C)^2}= R\;\; (Since |X _C|=20\;ohms)
    Z = 20 ohms
    I = 200 / 20 = 10 A
    V C = –JIXC.

  15. In a RC circuit consisting R = 10KΩ &phase angle between current &voltage is 45° What is the value of C at Ω = 1 ?

    &theata; = tan - 1 (XC /R)
    So, XC/R = 1
    XC = 10 KΩ
    or, 1/ C = 10KΩ.

  16. In a series RLC circuit the phasor diagram shown as below. Which of following is true regarding phasor diagram for the series RLC circuit ?

    Here VC > VL, 1/Ω C > Ω L.

  17. In a RLC series circuit consisting VR = 3V, V VC=8V & V VL = 4 V, Find the value of source excitation voltage ?

    V = \sqrt{(V_C_V_L)^2 + V_R^2} .

  18. In a parallel RL Circuit has i = 10A *amp; i VR = 8A. Find the value of current flowing through the inductor ?

    i_L = \sqrt {i^2 – i_R^2}.

  19. If a capacitor is added in parallel with inductor C in the above question, which draws a current of 6A then i will be

    i = \sqrt {i_R^2 + (i_C^2 _ i_L^2)^2} .

  20. Consider the following circuit diagram. If R = 25Ω X VL = X VC = 25Ω then what is the value of &theata; j ? [Assuming V as reference phasor]

    tan&theata; = X VL/R.

Please give us your valuable comment/suggestion. This will help us to improve this page.
Your Name : -

Your Email : - (Please do not forget to provide your actual email address so that we can send you the latest updates of this site. If you wish.)

Your Place : -

Your Comments and Suggestion : -



Enter the Security Code : -