An inductor and a resistor are connected in series to an ac source of voltage $144 \sin \left(100 \pi…

An inductor and a resistor are connected in series to an ac source of voltage $144 \sin \left(100 \pi t+\frac{\pi}{2}\right)$ Volt. If the current in the circuit is $6 \sin \left(100 \pi t+\frac{\pi}{6}\right)$ ampere, then the resistance of the resistor is
  1. $24 \Omega$
  2. $36 \Omega$
  3. $12 \Omega$
  4. $18 \Omega$

Solution

In LR circuit, $\begin{aligned} & V=144 \sin \left(100 \pi t+\frac{\pi}{2}\right) \\ & I=6 \sin \left(100 \pi t+\frac{\pi}{6}\right) \end{aligned}$
Also, $\tan \phi=\frac{X_L}{R} \Rightarrow \tan \left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\frac{X_L}{R}$ $\Rightarrow \sqrt{3}=\frac{X_L}{R} \Rightarrow X_L=\sqrt{3} R$ Now, $I_0=\frac{V_0}{z} \Rightarrow 6=\frac{144}{\sqrt{R^2+X_L^2}}=\frac{144}{\sqrt{R^2+(\sqrt{3} R)^2}}$ $\therefore \quad \mathrm{R}=12 \Omega$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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