Q19Basic Electrical Engineering

We are given the instantaneous equations for current and voltage in a series RL circuit: $i(t)=5\sin (314t+2\pi /3)$ $v(t)=15\sin (314t+5\pi /6)$

From these standard forms ($v=V_m\sin (\omega t+{\theta }_v)$), we can extract the peak values and phase angles: - $V_m=15\text{ V}$, phase angle ${\theta }_v=5\pi /6=150^{\circ }$ - $I_m=5\text{ A}$, phase angle ${\theta }_i=2\pi /3=120^{\circ }$ - Angular frequency $\omega =314\text{ rad/sec}$

(a) Impedance ($Z$): Impedance is the ratio of maximum voltage to maximum current. $Z=\frac{V_m}{I_m}=\frac{15}{5}=3\Omega$.

(e) Power Factor ($pf$): The phase difference ($\varphi$) between voltage and current is: $\varphi ={\theta }_v-{\theta }_i=150^{\circ }-120^{\circ }=30^{\circ }$. Since voltage leads current (expected in an RL circuit), the power factor is lagging. $pf=\cos (\varphi )=\cos (30^{\circ })=\frac{\sqrt{3}}{2}\approx 0.866\text{ lagging}$.

(b) Value of Resistance ($R$): Using the impedance triangle relationship, $R=Z\cos (\varphi )$. $R=3\times \cos (30^{\circ })=3\times 0.866=2.598\Omega$.

(c) Value of Inductance in Henrys ($L$): First, find the Inductive Reactance ($X_L$). $X_L=Z\sin (\varphi )$. $X_L=3\times \sin (30^{\circ })=3\times 0.5=1.5\Omega$. Since $X_L=\omega L$, we can solve for $L$: $L=\frac{X_L}{\omega }=\frac{1.5}{314}\approx 0.00477\text{ Henrys (or 4.77 mH)}$.

(d) Average Power Drawn ($P_{avg}$): Average real power consumed is calculated using RMS values: $P=V_{rms}{I}_{rms}\cos (\varphi )$. $V_{rms}=\frac{15}{\sqrt{2}}$ and $I_{rms}=\frac{5}{\sqrt{2}}$. $P=\left(\frac{15}{\sqrt{2}}\right)\times \left(\frac{5}{\sqrt{2}}\right)\times 0.866=\frac{75}{2}\times 0.866=37.5\times 0.866\approx 32.47\text{ Watts}$.