The frequency of an alternating voltage is 50 | vedclass

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(B) The instantaneous voltage $V$ is given by $V = V_0 \sin(\omega t)$,where $V_0$ is the peak voltage and $\omega = 2\pi f$.Given $f = 50 \ Hz$,so $\

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$\frac{1}{600} \ s$

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(B) The instantaneous voltage $V$ is given by $V = V_0 \sin(\omega t)$,where $V_0$ is the peak voltage and $\omega = 2\pi f$.Given $f = 50 \ Hz$,so $\omega = 2 	imes \pi 	imes 50 = 100\pi \ rad/s$.We want to find the time $t$ when $V = \frac{V_0}{2}$.Substituting this into the equation: $\frac{V_0}{2} = V_0 \sin(100\pi t)$.$\sin(100\pi t) = \frac{1}{2}$.Since $\sin(30^\circ) = \frac{1}{2}$,we have $100\pi t = \frac{\pi}{6}$.$t = \frac{\pi}{6 	imes 100\pi} = \frac{1}{600} \ s$.

$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2 \sqrt{2}}$

The frequency of an alternating voltage is $50 \ Hz$. The time taken for the instantaneous voltage to increase from zero to half of its peak voltage is:

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