Alternating Current MCQs for NEET — Physics Questions with Answers

Here, phase difference

           $\tan \varphi =\frac{X_L-X_C}{R}$

           $\tan \frac{\mathrm{\pi }}{3}=\frac{X_L-X_C}{R}$

When, L is removed

            $\sqrt{3}=\frac{X_C}{R}$

            $X_C=\sqrt{3}R$

When C is removed

          $tan\frac{\mathrm{\pi }}{3}=\sqrt{3}=\frac{X_L}{R}$

           $X_L=R\sqrt{3}$

Hence, in resonant circuit

       $\tan \varphi =\frac{\sqrt{3}R-\sqrt{3}R}{R}=0$

            $\varphi =0$

$\therefore Power factor \cos \varphi =1$

It is the condition of resonance therefore phase difference between voltage and current is zero and power factor is cos$\varphi$=1.

Given equations

            $\mathrm{i}=\frac{1}{\sqrt{2}}\sin \left(100\mathrm{\pi t}\right)$

and      $$\begin{gathered} \mathrm{e}=\frac{1}{\sqrt{2}}\sin (100\mathrm{\pi t}+\mathrm{\pi }/3) \\ \mathrm{so}, {\mathrm{i}}_{\mathrm{e}}=\frac{1}{\sqrt{2}}\mathrm{and} {\mathrm{V}}_{\mathrm{e}}=\frac{1}{\sqrt{2}} \end{gathered}$$

We know that average power

          $$\begin{gathered} {\mathrm{P}}_{\mathrm{av}}={\mathrm{V}}_{\mathrm{rms}}\times {\mathrm{i}}_{\mathrm{rms}} \mathrm{cos\varphi } \\ =\frac{1}{2}\times \frac{1}{2}\times \cos 60° \end{gathered}$$

                       $\left[\mathrm{since}, {\mathrm{i}}_{\mathrm{rms}}=\frac{{\mathrm{i}}_{\mathrm{o}}}{\sqrt{2}}\mathrm{and} {\mathrm{V}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_{\mathrm{o}}}{\sqrt{2}}\right]$

           $$\begin{gathered} =\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} \\ =\frac{1}{8}\mathrm{W} \end{gathered}$$

$$\begin{gathered} \mathrm{Given}, \mathrm{e}=200\sqrt{2}\sin 100\mathrm{t} \\ \mathrm{and} \mathrm{C}=1 \mathrm{\mu F} \end{gathered}$$

        $$\begin{gathered} {\mathrm{E}}_{\mathrm{rms}}=200 \mathrm{V} \\ {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\mathrm{\omega C}}=\frac{1}{1\times {10}^{-6}\times 100}={10}^4 \mathrm{\Omega } \\ {\mathrm{i}}_{\mathrm{rms}}=\frac{{\mathrm{E}}_{\mathrm{rms}}}{{\mathrm{X}}_{\mathrm{C}}·} \\ {\mathrm{i}}_{\mathrm{rms}}=\frac{200}{{10}^4}=2\times {10}^{-2} \mathrm{A}=20 \mathrm{mA} \end{gathered}$$

$$\begin{gathered} \tan \mathrm{\varphi }=\frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}=\frac{\mathrm{L\omega }}{\mathrm{R}} \\ \tan \mathrm{\varphi }=\frac{3\mathrm{\Omega }}{3\mathrm{\Omega }} \\ \tan \mathrm{\varphi }=1 \\ \mathrm{\varphi }={\tan }^{-1}\left(1\right) \\ \mathrm{\varphi }=45° \\ \mathrm{\varphi }=\frac{\mathrm{\pi }}{4}\mathrm{rad} \end{gathered}$$

Power dissipated in series L-C-R.

$P=I_{rms}^{2}R=\frac{{\epsilon }_{rms}^{2}R}{{\left|Z\right|}^2}=\frac{{\epsilon }^{2}R}{\left[R^2+{\left(\omega L-{\displaystyle \frac{1}{\omega C}}\right)}^2\right]}$

The power is defined as the rate at which work is being done in the circuit. 

Power= rate of work done in one complete cycle. 

or       $P_{av}=\frac{W}{T}$

or      $P_{av}=\frac{\left(E_o{I}_o\cos \varphi \right)T/2}{T}$

or      $P_{av}=\frac{E_o{I}_o}{2}\varphi$

where cos $\varphi$ is called the power factor of an AC circuit. 

The power factor of an L-C-R circuit is given by cosφ, where φ is the phase angle between voltage and current. The given potential differences across R, C, and L are 80V, 40V, and 100V respectively. Using these values, cosφ = 80/√(80^2 + 40^2 + 100^2) = 0.8. Therefore, the power factor is 0.8.

impedance of the R-C circuit, Z 
= $\sqrt{R^2+X_C^2}$ where, R =100$\Omega$ and X_C = 100$\Omega$ 
 
= 100$\sqrt{2}\Omega$ 
The peak value of the current, 
$I_{max}=\frac{V_{max}}{Z}=\frac{220\sqrt{2}}{100\sqrt{2}}=2.2 A$

In a series R-C circuit, the voltage drop across the capacitor is inversely proportional to the capacitance. When the capacitor is mica-filled, its capacitance increases due to the higher dielectric constant of mica compared to air. Therefore, the voltage drop across the mica-filled capacitor will be lower than the air-filled capacitor for the same current.