An alternating current I =14 sin (100 π t ) A passes through a series combination of a resistor of 30 Ω and an inductor of (2/(5 π)) H. Taking √ 2…

\(\displaystyle I=14 \sin (100 \pi t) \mathrm{~A}\) \(\displaystyle I_{rms}=\frac{I_{o}}{\sqrt{2}}=\frac{14}{1.4} \mathrm{~A}\) \(\displaystyle I_{rms}=10 \mathrm{~A}\) \(\displaystyle R=30 \Omega\) \(\displaystyle V_{R}=10 \times 30=300 \mathrm{~V}\) \(\displaystyle X_{L}=\omega L=100 \pi \times \frac{2}{5 \pi}=40 \Omega \quad\left[\therefore L=\frac{2}{5 \pi}\right]\) \(\displaystyle V_{L}=10 \times X_{L}=10 \times 40=400 \mathrm{~V}\) \(\displaystyle V_{L}=400 \mathrm{~V}\) \(\displaystyle Z=\sqrt{X_{L}^{2}+R^{2}}\) \(\displaystyle Z=\sqrt{(40)^{2}+(30)^{2}}\) \(\displaystyle =50 \Omega\) \(\displaystyle \cos \phi=\frac{30}{50}\) \(\displaystyle \cos \phi=\frac{3}{5}\)