 Irvin Dukes

2021-12-31

(a) Compute the reactance of a 0.450-H inductor at frequencies of 60.0 Hz and 600 Hz. (b) Compute the reactance of a $2.50-\mu$ capacitor at the same frequencies. (c) At what frequency is the reactance of a 0.450-H inductor equal to that of a $2.50-\mu$ capacitor? Alex Sheppard

Expert

Step 1
Given
The inductance of the coil is $L=0.45H$.
Solution
(a) For ac source connected with a coil, the inductance reactance ${X}_{L}$ in the circuit is given by equation 31.12 in the form
${X}_{L}=2\pi fL$
This inductance reactance is related to the frequency f and the inductance L of the coil. Now, we want to get ${X}_{L}$ at $f=60Hz$ and 600 Hz.
For $f=60Hz$:
${X}_{L}=2\pi fL=2\pi \left(60Hz\right)\left(0.45H\right)=170\mathrm{\Omega }$
For $f=600Hz$ :
${X}_{L}=2\pi fL=2\pi \left(600Hz\right)\left(0.45H\right)=1700\mathrm{\Omega }$ Expert

Step 2
(b) The capacitance of the capacitor is $C=2.5\mu F$. The capacitive reactance ${X}_{C}$ in the circuit is given by equation 31.18 in the form
${X}_{C}=\frac{1}{2\pi fC}$
Now, we calcaute ${X}_{C}$ or 60 Hz and 600 Hz as next
${X}_{C\left(60\right)}=\frac{1}{2\pi fC}=\frac{1}{2\pi \left(60Hz\right)\left(2.5×{10}^{-6}F\right)}=1.06×{10}^{3}\mathrm{\Omega }$
${X}_{C\left(600\right)}=\frac{1}{2\pi fC}=\frac{1}{2\pi \left(600Hz\right)\left(2.5×{10}^{-6}F\right)}=106\mathrm{\Omega }$ Vasquez

Expert

Step 3
(c) In this part, the two reactance are the same but at a specific frequency f. We are asked to determine the value of this frequency.
${X}_{C}={X}_{L}$
$\frac{1}{2\pi fC}=2\pi fL$
$f=\frac{1}{2\pi \sqrt{LC}}$
Now plug the values for LL and CC into equation (3) to get f
$=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{\left(0.45H\right)\left(2.5×{10}^{-6}F\right)}}=150Hz$

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