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(B) Given: $i = 100 \sin(314t) \ A$ and $e = 200 \sin(314t + \pi/3) \ V$.Comparing with standard equations $i = i_0 \sin(\omega t)$ and $e = e_0 \sin(

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$\sqrt{3} \ \Omega $

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(B) Given: $i = 100 \sin(314t) \ A$ and $e = 200 \sin(314t + \pi/3) \ V$.Comparing with standard equations $i = i_0 \sin(\omega t)$ and $e = e_0 \sin(\omega t + \phi)$,we get peak current $i_0 = 100 \ A$ and peak voltage $e_0 = 200 \ V$.The impedance $Z$ of the circuit is given by $Z = e_0 / i_0 = 200 / 100 = 2 \ \Omega$.We know that the impedance is related to resistance $R$ and reactance $X$ by the formula $Z^2 = R^2 + X^2$.Given $R = 1 \ \Omega$,we have $2^2 = 1^2 + X^2$.$4 = 1 + X^2 \implies X^2 = 3$.Therefore,the reactance $X = \sqrt{3} \ \Omega$.

The instantaneous values of current and voltage in an $AC$ circuit are $i = 100 \sin(314t) \ A$ and $e = 200 \sin(314t + \pi/3) \ V$ respectively. If the resistance is $1 \ \Omega$,then the reactance of the circuit will be:

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