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Question

(C) The given voltage equation is $V = 240 \sin(120t)$.Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the peak voltage $V_m = 2

Vedclass · Accepted Answer

$19 	ext{ Hz and } 170 	ext{ V}$

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(C) The given voltage equation is $V = 240 \sin(120t)$.Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the peak voltage $V_m = 240 	ext{ V}$ and angular frequency $\omega = 120 	ext{ rad/s}$.The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{120}{2 	imes 3.14159} \approx 19.1 	ext{ Hz}$.The $r.m.s.$ voltage is given by $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{240}{1.414} \approx 169.7 	ext{ V} \approx 170 	ext{ V}$.Thus,the frequency is $19 	ext{ Hz}$ and the $r.m.s.$ voltage is $170 	ext{ V}$.

Currents	$r.m.s.$ values
$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2\sqrt{2}}$

$A$ generator produces a voltage that is given by $V = 240 \sin(120t)$,where $t$ is in seconds. The frequency and $r.m.s.$ voltage are:

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