The peak value of an alternating current is 5A and its frequency is 60Hz. Find its rms value. How long will the cukrgrenQt current IS 5 A and its frequency is 60 Hz. Find its runs value.
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If the peak value of a current in 50Hz AC. Cirucit is 7.07 A. What is the mean value of current over half a cycle and the value of current `1//300`s a
As `I_(m) = (2I_(0))/pi`, where `I_(0)` = peak value of current = 7.07 A `therefore I_(m) = 0.637 xx 7.07=4.5A` Also, at t=`(1/300)`s `I=I_(0)sinomegat = I_(0) sin 2pift` `(therefore omega=2pif)`...
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a) The peak voltage of an AC supply is 300 V. What is the rms voltage? b) The rms value of current in an AC circuit is 10A. What is the peak current?
a) Heak, peak value of AC supply, `V_(0) = 300V` `therefore` Using the relation `V_(rms) = V_(0)/sqrt(2)` We get, `V_(rms) = 300/sqrt(2) = 212.1`V b) Let `I_(0)` be the peak...
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The peak value of an alternating current is `5 A` and its frequency is `60 Hz`. Find its rms value. How long will the current take to reach the peak v
Correct Answer - A `underset(rms)(I)` = `underset(o)(I)`/ `sqrt 2` = 5/`sqrt 2` = 3.536 A Time taken, T = 1/`4 xx 60` = 4.167ms
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An alternating current in a circuit is given by l = 20 sin ( 100`pi`t + ` 0.05pi `) A. The rms value and the frequency of current respectively are
Correct Answer - C I = 20 sin(100`pi`t + `0,05pi`) `therefore` `underset(rms)(I)` = 20 / `sqrt 2 ` = 10`sqrt 2` A `omega` = `100pi` `adr` f = 50Hz
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The peak value of an alternating emf E given by E = `underset(o)(E) ` cos `omega`t is 10 V and frequency is 50 Hz . At time t = (1/600) s, the instant
Correct Answer - B E = 10 cos(`2pi`ft) = 10 cos (`2pi xx 50 xx 1/600`) = 10 cos`pi`/6 = `5sqrt 3`V
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In a heating arrangement , an alternating current having a peak value of 28 A is used . To produce the same heat energy , If the constant current is u
Correct Answer - C `underset(o)(I)` = 28 A `therefore` `underset(rms)(I)` = `underset(o)(I)`/`sqrt 2` `approx` 20 A
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The alternating current in a circuit is given by `I = 50sin314t`. The peak value and frequency of the current are
Correct Answer - B From standard equation, we have `I= I_(0)sinomegat`...............(i) Given, I=50sin314t ..............(ii) Comparing Eqs. (i) and (ii), we get `I_(0)`= 50 A, `omega = 2pif` = 314 `rArr f=314/(2...
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Assertion The average value of alternating emf is 63.39% of the peak value . Reason The rms value of alternating emf is70.72% of peak value
Correct Answer - B The average value of alternating emf `E=E_(0)sinomega` t over a positive half cycle is `E_(av) half cycle =(2E_(0))/(pi)=0.6369` Percentage of `E_(av)=0.6369xx100=63.69%` Also rms value of emf is...
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An inductor (L=2000mH) is connected to an AC source of peak current. What is the instantaneous voltage of the source when the current is at its peak v
Because phase difference between voltage and current is `pi//2` for pure inductor. So, Ans is zero
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A long solenoid having 1000 turns per cm carries an alternating current of peak value 1 A. A search coil having a cross-sectional area of `1xx10^(-4)
Correct Answer - `15.8 s^(-1)`
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