(C) The given equation is $V = V_0 \sin(\omega t)$,where $V_0 = 311 \ V$ and $\omega = 314 \ rad/s$.The $rms$ value of voltage is given by $V_{rms} =

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(C) The given equation is $V = V_0 \sin(\omega t)$,where $V_0 = 311 \ V$ and $\omega = 314 \ rad/s$.The $rms$ value of voltage is given by $V_{rms} = \frac{V_0}{\sqrt{2}}$.$V_{rms} = \frac{311}{1.414} \approx 220 \ V$.The angular frequency is $\omega = 2\pi f$.$314 = 2 \times 3.14 \times f$.$314 = 6.28 \times f$.$f = \frac{314}{6.28} = 50 \ Hz$.Thus,the $rms$ voltage is $220 \ V$ and the frequency is $50 \ Hz$.

Class 12 Physics Alternating Current Alternating Current, Voltage (rms and Average)

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$A$ sinusoidal voltage produced by an $AC$ generator at any instant $t$ is given by the equation $V = 311 \sin(314t)$. The $rms$ value of voltage and frequency are respectively:

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A
$200 \ V, 50 \ Hz$
B
$220 \ V, 100 \ Hz$
C
$220 \ V, 50 \ Hz$
D
$200 \ V, 100 \ Hz$

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Alternating Current, Voltage (rms and Average)

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$A$ sinusoidal voltage produced by an $AC$ generator at any instant $t$ is given by the equation $V = 311 \sin(314t)$. The $rms$ value of voltage and frequency are respectively:

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The output voltage of a step-down transformer is measured to be $24 \text{ V}$, when connected to a $12 \text{ W}$ light bulb. The value of the peak current is . . . . . . .

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