In an ac circuit,the instantaneous voltage e( | vedclass

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(D) Given voltage $e(t) = 5[\cos \omega t + \sqrt{3} \sin \omega t]$.Multiplying and dividing by $2$,we get $e(t) = 10[\frac{1}{2} \cos \omega t + \fr

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Current leads voltage by $\frac{\pi}{12}$

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(D) Given voltage $e(t) = 5[\cos \omega t + \sqrt{3} \sin \omega t]$.Multiplying and dividing by $2$,we get $e(t) = 10[\frac{1}{2} \cos \omega t + \frac{\sqrt{3}}{2} \sin \omega t]$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write $e(t) = 10 \sin(\omega t + \frac{\pi}{3})$.The current is given as $i(t) = 5 \sin(\omega t + \frac{\pi}{4})$.The phase of voltage is $\phi_v = \frac{\pi}{3}$ and the phase of current is $\phi_i = \frac{\pi}{4}$.The phase difference is $\Delta \phi = \phi_v - \phi_i = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$.Since $\phi_v > \phi_i$,the voltage leads the current by $\frac{\pi}{12}$. However,checking the options,the current leads the voltage by $-\frac{\pi}{12}$ or voltage leads current by $\frac{\pi}{12}$. Given the standard interpretation of such problems,if the current leads by $\frac{\pi}{12}$ is not an option,we re-evaluate: $\phi_i - \phi_v = \frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$. Thus,voltage leads current by $\frac{\pi}{12}$. Since this is not listed,let's re-check the calculation: $\frac{\pi}{3} \approx 60^\circ$,$\frac{\pi}{4} = 45^\circ$. $60^\circ - 45^\circ = 15^\circ = \frac{\pi}{12}$. The correct physical result is voltage leads current by $\frac{\pi}{12}$. Given the provided options,there might be a typo in the question's options,but based on the math,the phase difference is $\frac{\pi}{12}$.

In an $ac$ circuit,the instantaneous voltage $e(t)$ and current $i(t)$ are given by $e(t) = 5[\cos \omega t + \sqrt{3} \sin \omega t] \ V$ and $i(t) = 5[\sin(\omega t + \frac{\pi}{4})] \ A$. Determine the phase relationship between voltage and current.

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