What is the total inductive reactance of each circuit of Figure 13-33?

Question

Accepted Answer

The reactances of the individual inductors are the same in both circuits. From Equation 13-12 : [latex]X_{L}= 2 \pi fL[/latex],

[latex]X_{L1}= 2\pi fL_{1}= 2\pi (200 \ kHz)(2.7 \ mH)= 3.93 \ K\Omega [/latex]

[latex]X_{L2}= 2\pi fL_{2}= 2\pi (200 \ kHz)(4.7 \ mH)= 5.91 \ K\Omega [/latex]

For the series inductors in Figure 13-33(a), the total reactance is the sum of [latex]X_{L1}[/latex]and [latex]X_{L2}[/latex], as given in Equation 13-13 : [latex]X_{L(tot)}= X_{L1}+ X_{L2}+ X_{L3}+....+ X_{Ln}[/latex].

[latex]X_{L(tot)}= X_{L1}+ X_{L2}= 3.39 \ k\Omega + 5.91 \ k\Omega =9.30 \ k\Omega [/latex]

For the inductors in parallel in Figure 13-33(b), determine the total reactance by Equation 13-14 : [latex]X_{L(tot)}= \frac{1}{\frac{1}{X_{L1}}+ \frac{1}{X_{L2}} + \frac{1}{X_{L3}}+ .......+ \frac{1}{X_{Ln}}} [/latex] or from the produet-over-sum rule using [latex]X_{L1}[/latex]and [latex] X_{L2}[/latex]

[latex]X_{L(tot)}= \frac{X_{L1}X_{L2}}{X_{L1}+ X_{L2}} = \frac{(3.39 \ k\Omega)(5.91 \ k\Omega )}{3.39 \ k\Omega + 5.91 \ k\Omega} = 2.15 \ k\Omega [/latex]

You can also obtain the total reactance for either series or parallel inductors by first finding the total inductance and then substituting in Equation 13-12 : [latex]X_{L}= 2 \pi fL[/latex] to find the total reactance

Question 13.13: What is the total inductive reactance of each circuit of Fig...

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