Chapter 21, Problem 56P

To determine

The coil’s inductance

Answer to Problem 56P

Solution:

The coil induction will be $4.97\times {140}^{-2}\text{H}$

Explanation of Solution

Given info:

$f=10.0\times {10}^3$ Hz

$L=0.160$ H

$V=240$ V

Formula used:

$X_L=2\pi fL$

Where,

$f=10.0\times {10}^3$ Hz

$L=0.160$ H

$V=240$ V

Calculation:

$\begin{array}{l}2\pi fL=2\pi \left(10.0\times {10}^3\text{Hz}\right)\left(0.160\text{H}\right)\\ =10053\text{Ω}\approx 10.0\text{kΩ}\end{array}$

$X_L\to I=\frac{V}{X_L}=\frac{240\text{V}}{10053\text{Ω}}=2.39\times {10}^{-2}\text{A}$

We can find the coil induction by

$\begin{array}{l}V=I{X}_L\to X_L=\frac{V}{I}\\ X_L=2\pi fL\to L=\frac{X_L}{2\pi f}=\frac{\text{V}}{2\pi fI}=\frac{240\text{V}}{2\pi \left(60.0\text{Hz}\right)12.8\text{A}}=4.97\times {140}^{-2}\text{H}\end{array}$