In an LCR series circuit,an alternating e.m.f | vedclass

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(C) Given equations are $e = 160 \sin(100t) 	ext{ V}$ and $i = 250 \sin(100t + \frac{\pi}{3}) 	ext{ mA}$.Comparing these with standard forms $e = e_

Vedclass · Accepted Answer

$10$

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(C) Given equations are $e = 160 \sin(100t) 	ext{ V}$ and $i = 250 \sin(100t + \frac{\pi}{3}) 	ext{ mA}$.Comparing these with standard forms $e = e_0 \sin(\omega t)$ and $i = i_0 \sin(\omega t + \phi)$,we get:Peak voltage $e_0 = 160 	ext{ V}$.Peak current $i_0 = 250 	ext{ mA} = 250 	imes 10^{-3} 	ext{ A} = 0.25 	ext{ A}$.Phase difference $\phi = \frac{\pi}{3}$.The average power dissipated in an $AC$ circuit is given by $P_{avg} = V_{rms} I_{rms} \cos(\phi) = \frac{e_0}{\sqrt{2}} \cdot \frac{i_0}{\sqrt{2}} \cos(\phi) = \frac{e_0 i_0}{2} \cos(\phi)$.Substituting the values:$P_{avg} = \frac{160 	imes 0.25}{2} 	imes \cos(\frac{\pi}{3})$$P_{avg} = \frac{40}{2} 	imes \frac{1}{2} = 20 	imes 0.5 = 10 	ext{ W}$.

In an $LCR$ series circuit,an alternating e.m.f. '$e$' and current '$i$' are given by the equations $e = 160 \sin(100t) \text{ V}$ and $i = 250 \sin(100t + \frac{\pi}{3}) \text{ mA}$. The average power dissipated in the circuit is: (in $\text{ W}$)

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