The a.c. source of e.m.f. with instantaneous | vedclass

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(B) The instantaneous e.m.f. is given by $e = 200 \sin(50t) 	ext{ V}$.Comparing this with the standard equation $e = e_0 \sin(\omega t)$, we get the

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$2.828$

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(B) The instantaneous e.m.f. is given by $e = 200 \sin(50t) 	ext{ V}$.Comparing this with the standard equation $e = e_0 \sin(\omega t)$, we get the peak e.m.f. $e_0 = 200 	ext{ V}$.The peak current $I_0$ is calculated using Ohm's law: $I_0 = \frac{e_0}{R} = \frac{200}{50} = 4 	ext{ A}$.The r.m.s. value of current $I_{	ext{rms}}$ is given by the formula $I_{	ext{rms}} = \frac{I_0}{\sqrt{2}}$.Substituting the values: $I_{	ext{rms}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} 	ext{ A}$.Since $\sqrt{2} \approx 1.414$, we have $I_{	ext{rms}} = 2 	imes 1.414 = 2.828 	ext{ A}$.

The a.c. source of e.m.f. with instantaneous value $e$ is given by $e = 200 \sin(50t) \text{ V}$. The r.m.s. value of current in a circuit of resistance $50 \ \Omega$ is: (in $\text{ A}$)

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