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A dc voltage with ripple is given by V(t) = [100 + 10 sin (ωt) - 5 sin (3ωt)] volts. Measurements of this voltage v(t), made by moving-coil and moving-iron voltmeters, show, readings of V1 and V2 respectively. The value of  V2 - V1, in volts, is _________.
    Correct answer is between '0.30,0.33'. Can you explain this answer?
    Most Upvoted Answer
    A dc voltage with ripple is given by V(t) = [100 + 10 sin (ωt) -...
    Concept:
    PMMC measures dc or average value, hot-wire and moving iron instruments measures RMS value.
    If a current i is given as:
    i = a0 + a1 sin (ω1t + θ1) + a2 sin (ω2t + θ2) + a3 sin (ω3t + θ3) +...........
    Where,
    a0 = DC value = average value of current
    Rms value is:
    Calculation:
    Given,
    V(t) = [100 + 10 sin (ωt) - 5 sin (3ωt)] 
    In the above DC value of the signal is 100 V
    For moving coil,
    V1 = Vavg = 100 V
    For moving iron,

    The difference between the two meters is 
    V2 - V1 = 0.312 volts
    Free Test
    Community Answer
    A dc voltage with ripple is given by V(t) = [100 + 10 sin (ωt) -...
    Ωt)] volts, where ω is the angular frequency in rad/s and t is the time in seconds.

    The peak value of the ripple voltage is 10 volts, and the angular frequency ω determines the frequency of the ripple.

    To find the frequency of the ripple, we can use the formula:

    f = ω / (2π)

    where f is the frequency in Hz and ω is the angular frequency in rad/s.

    Given that the angular frequency ω is in rad/s, we can calculate the frequency of the ripple by dividing ω by 2π.

    f = ω / (2π) = 10 / (2π) ≈ 1.59 Hz

    Therefore, the frequency of the ripple is approximately 1.59 Hz.
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    A dc voltage with ripple is given by V(t) = [100 + 10 sin (ωt) - 5 sin (3ωt)] volts. Measurements of this voltage v(t), made by moving-coil and moving-iron voltmeters, show, readings of V1 and V2 respectively. The value of V2 - V1, in volts, is _________.Correct answer is between '0.30,0.33'. Can you explain this answer?
    Question Description
    A dc voltage with ripple is given by V(t) = [100 + 10 sin (ωt) - 5 sin (3ωt)] volts. Measurements of this voltage v(t), made by moving-coil and moving-iron voltmeters, show, readings of V1 and V2 respectively. The value of V2 - V1, in volts, is _________.Correct answer is between '0.30,0.33'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A dc voltage with ripple is given by V(t) = [100 + 10 sin (ωt) - 5 sin (3ωt)] volts. Measurements of this voltage v(t), made by moving-coil and moving-iron voltmeters, show, readings of V1 and V2 respectively. The value of V2 - V1, in volts, is _________.Correct answer is between '0.30,0.33'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dc voltage with ripple is given by V(t) = [100 + 10 sin (ωt) - 5 sin (3ωt)] volts. Measurements of this voltage v(t), made by moving-coil and moving-iron voltmeters, show, readings of V1 and V2 respectively. The value of V2 - V1, in volts, is _________.Correct answer is between '0.30,0.33'. Can you explain this answer?.
    Solutions for A dc voltage with ripple is given by V(t) = [100 + 10 sin (ωt) - 5 sin (3ωt)] volts. Measurements of this voltage v(t), made by moving-coil and moving-iron voltmeters, show, readings of V1 and V2 respectively. The value of V2 - V1, in volts, is _________.Correct answer is between '0.30,0.33'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
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