A resistor and an inductor are connected in series to an AC source of voltage $150 \sin (100 \pi t+\pi)$ volt. If the current in the circuit is $5 \sin \left(100 \pi t+\frac{2 \pi}{3}\right)$ ampere, then the average power dissipated and the resistance of the resistor are respectively

• A

  $187.5 \mathrm{~W}, 30 \Omega$
• Correct answer
  B

  $187.5 \mathrm{~W}, 15 \Omega$
• C

  $375 \mathrm{~W}, 30 \Omega$
• D

  $375 \mathrm{~W}, 15 \Omega$

Solution

Given, peak voltage, $V_0=150 \mathrm{~V}$ Peak current, $$ I_0=5 $$ Phase difference between voltage and current, $$ \phi=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}=60^{\circ} $$ Now, average power in $R-L$ circuit, $$ \begin{aligned} P_{\mathrm{av}}=V_{\mathrm{rms}} & \cdot I_{\mathrm{rms}} \cdot \cos \phi \\ & =\frac{V_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cos 60^{\circ}=\frac{150}{\sqrt{2}} \times \frac{5}{\sqrt{2}} \times \frac{1}{2} \\ P_{\mathrm{av}}= & 187.5 \mathrm{~W} \end{aligned} $$
$\begin{gathered}\text { Impendence, } Z=\frac{V_0}{I_0}=\frac{150}{5} \\ Z=30 \Omega \\ \frac{R}{Z}=\cos \phi \\ \frac{R}{30}=\cos 60^{\circ} \Rightarrow R=30 \times \frac{1}{2} \\ R=15 \Omega\end{gathered}$

Hence, the correct answer is option: 'B'.