In an A.C. circuit,the potential difference V | vedclass

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(C) Given equations are $V = 100 \sin(100t) 	ext{ V}$ and $I = 100 \sin(100t + \frac{\pi}{3}) 	ext{ mA}$.Comparing with standard forms $V = V_0 \sin

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$2.5 	ext{ W}$

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(C) Given equations are $V = 100 \sin(100t) 	ext{ V}$ and $I = 100 \sin(100t + \frac{\pi}{3}) 	ext{ mA}$.Comparing with standard forms $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$,we get:$V_0 = 100 	ext{ V}$$I_0 = 100 	ext{ mA} = 100 	imes 10^{-3} 	ext{ A} = 0.1 	ext{ A}$Phase difference $\phi = \frac{\pi}{3}$.The average power dissipated in an $A$.$C$. circuit is given by $P = V_{	ext{rms}} I_{	ext{rms}} \cos \phi$.$P = \left(\frac{V_0}{\sqrt{2}}ight) \left(\frac{I_0}{\sqrt{2}}ight) \cos \phi = \frac{V_0 I_0}{2} \cos \phi$.Substituting the values:$P = \frac{100 	imes 0.1}{2} 	imes \cos(\frac{\pi}{3})$$P = \frac{10}{2} 	imes \frac{1}{2} = 5 	imes 0.5 = 2.5 	ext{ W}$.

In an $A$.$C$. circuit,the potential difference $V$ and current $I$ are given respectively by $V = 100 \sin(100t) \text{ V}$ and $I = 100 \sin(100t + \frac{\pi}{3}) \text{ mA}$. The power dissipated in the circuit will be (Given: $\cos \frac{\pi}{3} = \frac{1}{2}$)

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In an $A.C.$ circuit,$V$ and $I$ are given by:
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$I = 100 \sin(100 t + \pi/3) \text{ mA}$
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