The equation of current in a purely inductive | vedclass

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(D) The given current equation is $i = i_0 \sin (\omega t - 30^{\circ})$,where $i_0 = 5 \, A$ and $\omega = 49 \pi \, rad/s$.In a purely inductive cir

Vedclass · Accepted Answer

$23.1 \sin (49 \pi t + 60^{\circ})$

Vedclass · Answer

(D) The given current equation is $i = i_0 \sin (\omega t - 30^{\circ})$,where $i_0 = 5 \, A$ and $\omega = 49 \pi \, rad/s$.In a purely inductive circuit,the inductive reactance is $X_L = \omega L$.Given $L = 30 \, mH = 30 	imes 10^{-3} \, H$.$X_L = (49 	imes \frac{22}{7}) 	imes 30 	imes 10^{-3} = (7 	imes 22) 	imes 30 	imes 10^{-3} = 154 	imes 30 	imes 10^{-3} = 4.62 \, \Omega$.The peak voltage is $v_0 = i_0 X_L = 5 	imes 4.62 = 23.1 \, V$.In a purely inductive circuit,the voltage leads the current by $90^{\circ}$.Therefore,the phase of the voltage is $(-30^{\circ} + 90^{\circ}) = +60^{\circ}$.The voltage equation is $v = v_0 \sin (\omega t + 60^{\circ}) = 23.1 \sin (49 \pi t + 60^{\circ})$.

The equation of current in a purely inductive circuit is $i = 5 \sin (49 \pi t - 30^{\circ})$. If the inductance is $30 \, mH$,then the equation for the voltage across the inductor will be. (Let $\pi = \frac{22}{7}$)

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