Question

An inductance of (200)/(pi) mH a capacitance of (10^(-3))/(pi) and a resistance of 10 Omega are connected in series with an AC source of 220 V, 50Hz. The phase angle of the circuit is

Answer 1

Given:

• Inductance $(L) = \frac{200}{\pi} mH$

• Capacitance $(C) = \frac{10^{-3}}{\pi} F$

• Resistance $(R) = 10 \Omega$

• Voltage $(V) = 220V$

• Frequency $(n) = 50Hz$

To Find:

• Phase angle of the circuit

Solution:

Here,

          $L = \frac{200}{\pi} mH$

          $L = \frac{200 \times 10^{-3}}{\pi} H$

          $L = \frac{0.2}{H} H$

Inductance of a circuit is given by

⇒      $X_L = \omega L$

         $X_L = 2 \pi n L$

        $X_L = 2 \pi \times 50 \times \frac{0.2}{\pi} = 20\Omega$

Capacitance of a circuit is given by

⇒     $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi n C}$

       $X_C = \frac{\pi}{2 \pi \times 50 \times 10^{-3}} =10\Omega$

$tan \phi = \frac{(X_L-X_C)}{R}$

$tan \phi=\frac{20-10}{10}=1$

  $\phi = tan^{-1}(1)$

  $\phi = \frac{\pi}{4}$

Thus, the phase angle is  $\phi = \frac{\pi}{4}$