A series RLC circuit is driven by a generator at a frequency of$\mathbf{2000}\mathbf{}\mathbf{Hz}$and an emf amplitude of $\mathbf{170}\mathbf{}\mathbf{V}$. The inductance is$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mH}$ , the capacitance is$\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{}\mathbf{\mu F}$ , and the resistance is $\mathbf{200}\mathbf{}\mathbf{\Omega }$. (a) What is the phase constant in radians? (b) What is the current amplitude?

Frequency is $f=2000\mathrm{Hz}$

Emf amplitude is $\epsilon =170\mathrm{V}$.

Inductance is $L=60.0\mathrm{mH} or 60.0\times {10}^{-3}\text{H}$

Capacitance is $C=0.400\mathrm{\mu F} \mathrm{or} 0.400\times {10}^{-6}\mathrm{F}$

Reactance withstands currents without loss of power, unlike resistors. Inductive reactance increases with frequency and inductance. Capacitive reactance decreases with frequency and capacitance. Impedance represents the total opposition provided by reactance and resistance.

Formulas:

Inductive reactance is,

$X_L=2\pi fL$

Capacitive reactance is,

$X_c=\frac{1}{2\pi fC}$

The impedance is,

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

The phase angle is,
$\varphi ={\tan }^{-1}\left(\frac{X_L-X_c}{R}\right)$

The impedance is,

$Z=\frac{\text{emf} amplitude}{i}$

Here, $i$is the current amplitude.

Determine the inductive reactance as below.

$\begin{array}{rcl}{X}_L& =& 2\pi fL\\ & =& 2\times 3.14\times 2000\times 60.0\times {10}^{-3}\\ & =& 753.9822\Omega \end{array}$

Define the capacitive reactance is,

$\begin{array}{rcl}{X}_c& =& \frac{1}{2\pi fC}\\ & =& \frac{1}{2\times 3.14\times 2000\times 0.4\times {10}^{-6}}\\ & =& 199\Omega \end{array}$

The phase angle is,

$\begin{array}{rcl}\varphi & =& {\tan }^{-1}\left(\frac{X_L-X_c}{R}\right)\\ & =& {\tan }^{-1}\left(\frac{753.98-199}{200}\right)\\ & =& 70.18°\\ & =& 1.22\mathrm{rad}\end{array}$

Here, phase constant in radian is $1.22\mathrm{rad}$.

First define the impedance as below.

$\begin{array}{rcl}Z& =& \sqrt{R^2+{\left(X_L-X_C\right)}^2}\\ & =& \sqrt{{200}^2+\left({753.98}^2-{199}^2\right)}\\ & =& 598.56\Omega \end{array}$

Now calculate the current amplitude as below.

$\begin{array}{rcl}z& =& \frac{emf amplitude}{i}\\ 598.56& =& \frac{170}{i}\end{array}$

$i=0.288\mathrm{A}$

Hence, the current amplitude is $0.288\mathrm{A}$.