Chapter 21, Problem 87GP

To determine

To find the capacitor C whose resistance is $15.8\Omega$ and the current of the voltage are to be in phase.

Answer to Problem 87GP

Solution:

The rounding off to two significant digits the value of capacitance is $0.1054\mu F.$

Explanation of Solution

Given info:

$\omega$ Is the angular frequency and L is the inductance.

Formula used:

$X_L=\omega L$

Calculation:

The expression for reactance of an inductor is given as

$X_L=\omega L\mathrm{............}\left(1\right)$

$\begin{array}{l}{f}_0=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}\\ f_0{}^2=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}\\ f_0{}^2=\frac{1}{4{\pi }^{2}LC}\\ C=\frac{1}{4{\pi }^{2}LC}\\ C=\frac{1}{4{\pi }^{2}L{f}_0{}^2}\\ L=130mH\\ =130mH\left(\frac{{10}^{-3}H}{1mH}\right)\\ =130\times {10}^{-3}H\\ C=\frac{1}{4{\pi }^2{f}_0{}^{2}L}\\ C=\frac{1}{4{\pi }^2{\left(1360HZ\right)}^2\left(130\times {10}^{-3}H\right)}\\ =0.1054\times {10}^{-6}F\left(\frac{1\mu F}{{10}^{-6}F}\right)\\ =0.1054\mu F\end{array}$