Question

A $0.21-H$ inductor and a $88-\Omega$ resistor are connected in series to a $220-V,50-Hz$ AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. Use $\pi =22/7$

• A. $2A,ta{n}^{-1}3/4$
• B. $14.4A,ta{n}^{-1}7/8$
• C. $14.4A,ta{n}^{-1}8/7$
• D. $3.28A,ta{n}^{-1}2/11$

Answer 1

The correct option is A $2A,ta{n}^{-1}3/4$

The impedance of the circuit is given as,

$Z=\sqrt{R^2+{X_L}^2}$

$Z=\sqrt{{\left(88\right)}^2+{\left(2\times \pi \times 50\times 0.21\right)}^2}$

$Z=110\Omega$

The current in the circuit is given as,

$I=\frac{V}{Z}$

$I=\frac{220}{110}$

$=2A$

The phase angle is given as,

$\theta ={\tan }^{-1}\left(\frac{X_L}{R}\right)$

$\theta ={\tan }^{-1}\left(\frac{2\times \pi \times 50\times 0.21}{88}\right)$

$={\tan }^{-1}\left(\frac{3}{4}\right)$

Thus, the current in the circuit is $2A$ and the phase angle is ${\tan }^{-1}\left(\frac{3}{4}\right)$.