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

Beginner2022-01-01Added 36 answers

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$f=600Hz$ :

$X}_{L}=2\pi fL=2\pi \left(600Hz\right)\left(0.45H\right)=1700\mathrm{\Omega$

Given

The inductance of the coil is

Solution

(a) For ac source connected with a coil, the inductance reactance

This inductance reactance is related to the frequency f and the inductance L of the coil. Now, we want to get

For

For

Pademagk71

Beginner2022-01-02Added 34 answers

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

$X}_{C\left(60\right)}=\frac{1}{2\pi fC}=\frac{1}{2\pi \left(60Hz\right)(2.5\times {10}^{-6}F)}=1.06\times {10}^{3}\mathrm{\Omega$

$X}_{C\left(600\right)}=\frac{1}{2\pi fC}=\frac{1}{2\pi \left(600Hz\right)(2.5\times {10}^{-6}F)}=106\mathrm{\Omega$

(b) The capacitance of the capacitor is

Now, we calcaute

Vasquez

Expert2022-01-07Added 669 answers

Step 3

(c) In this part, the two reactance

Now plug the values for LL and CC into equation (3) to get f

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