A generator of frequency $\mathbf{3000}\mathbf{}\mathbf{Hz}$drives a series RLC circuit with an emf amplitude of$\mathbf{120}\mathbf{}\mathbf{V}$.The resistance islocalid="1662984209739" $\mathbf{40}\mathbf{}\mathbf{Ohm}$, the capacitance is$\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{\mu F}$, and the inductance is$\mathbf{850}\mathbf{}\mathbf{\mu H}$. What are (a) the phase constant in radians and (b) the current amplitude? (c) Is the circuit capacitive, inductive, or in resonance?

Frequency is $3000\mathrm{Hz}$.

Emf voltage is$120\mathrm{V}$.

Resistance is$40\mathrm{Ohm}$.

Capacitance is$1.60\mathrm{\mu F}$.

Inductance is $850\mathrm{\mu H}$.

Reactance withstand 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

$\begin{array}{rcl}{X}_L& =& 2\pi fL\\ & =& 2\times 3.14\times 3000\times 850\times {10}^{-6}\\ & =& 16\Omega \end{array}$

Define the capacitive reactance is,

$\begin{array}{rcl}{X}_c& =& \frac{1}{2\pi fC}\\ & =& \frac{1}{2\times 3.14\times 3000\times 1.60\times {10}^{-6}}\\ & =& 33.1\Omega \end{array}$

Now impedance is as follows

$\begin{array}{rcl}Z& =& \sqrt{R^2+{\left(X_L-X_C\right)}^2}\\ & =& \sqrt{{40}^2+{\left(16-33.1\right)}^2}\\ & =& 43.5\Omega \end{array}$

The phase angle is define by,

$\begin{array}{rcl}\varphi & =& {\tan }^{-1}\left(\frac{X_L-X_c}{R}\right)\\ & =& {\tan }^{-1}\left(\frac{16-33.1}{40}\right)\\ & =& -23.15°\\ & =& -0.404\mathrm{rad}\end{array}$

Phase constant is$-0.404\mathrm{rad}$

Write the emf in terms of impedance and current as below

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

Substitute known values in the above equation

$\begin{array}{rcl}i& =& \frac{120}{43.5}\\ & =& 2.76A\end{array}$

Hence, the current amplitude is$2.76\mathrm{A}$

Here the phase angle is negative, so the circuit is capacitive.