Chapter 21, Problem 64P

To determine

(a) To determine:

The rms current is the rms voltage divided by the impedance

Answer to Problem 64P

Solution:

The rms current is the rms voltage divided by the impedance will be $4.3\times {10}^{-2}A$

Explanation of Solution

Given:

$\begin{array}{l}f=60\text{Hz}\\ L=350\text{mH}\\ R=2.80\text{kΩ}\end{array}$

Formula Used:

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

Where;

$f=$ Frequency

$R=$ Resistance

$L=$ Inductor

The rms current is the rms voltage divided by the impedance

$I_{rms}=\frac{V_{rms}}{Z}=\frac{V_{rms}}{\sqrt{R^2+4{\pi }^2{f}^2{L}^2}}$

$\frac{V_{rms}}{\sqrt{R^2+4{\pi }^2{f}^2{L}^2}}$$=\frac{120V}{\sqrt{{\left(2.80\times {10}^3\Omega \right)}^2+4{\pi }^2{\left(60.0Hz\right)}^2{\left(0.35H\right)}^2}}$

$\begin{array}{l}=\frac{120V}{2803\Omega }\\ =0.04281\\ =4.3\times {10}^{-2}A\end{array}$

To determine

(b) To determine:

The phase angle

Answer to Problem 64P

Solution:

The phase angle will be ${2.7}^0$

Explanation of Solution

The phase angle

$\varphi ={\tan }^{-1}\frac{X_L}{R}={\tan }^{-1}\frac{2\pi fL}{R}=\frac{2\pi \left(60.0Hz\right)\left(0.35H\right)}{2.80\times {10}^3\Omega }={2.7}^0$

To determine

(c) To determine:

The power dissipated

Answer to Problem 64P

Solution:

The power dissipated will be $5.1\text{W}$

Explanation of Solution

The power dissipated

$P=I_{rms}^{2}R={\left(4.281\times {10}^{-2}A\right)}^2\left(2.80\times {10}^3\Omega \right)=5.1\text{W}$

To determine

(d) To determine:

The rms voltage reading

Answer to Problem 64P

Solution:

The rms voltage reading will be $120V\text{}$ and $5.6V$

Explanation of Solution

The rms voltage reading

$V_R=I_{rms}R=\left(4.281\times {10}^{-2}A\right)\left(2.80\times {10}^3\Omega \right)=119.87V=120V$

$V_L=I_{rms}{X}_L=I_{rms}2\pi fL=\left(4.281\times {10}^{-2}A\right)2\pi \left(60.0Hz\right)\left(0.35H\right)=5.649V=5.6V$