What is the total capacitive reactance of each circuit in Figure 12-45?

Question

Accepted Answer

The reactances of the individual capacitors are the same in both circuits.

[latex]X_{C1} = \frac{1}{2\pi fC_{1}} = \frac{1}{2\pi (5.0 \ kHz)(0.01 \ \mu F)}= 3.18 \ k\Omega[/latex]

[latex]X_{C2} = \frac{1}{2\pi fC_{2}} = \frac{1}{2\pi (5.0 \ kHz)(0.068 \ \mu F)}= 468 \ \Omega[/latex]

The Series Circuit: For the capacitors in series in Figure 12-45 (a), the total reactance is the sum of [latex]X_{C1}[/latex] and [latex]X_{C2}[/latex] , as given in Equation 12-26 : [latex]X_{C(tot)} = X_{C1}+ X_{C2}+ X_{C3}+ .....+ X_{Cn} [/latex].

[latex]X_{C(tot)}[/latex] = [latex]X_{C1}[/latex] + [latex] X_{C2}[/latex] = 3.18 kΩ + 468 Ω = 3.65 kΩ

Alternatively, you can obtain the total series reactance by first finding the total capacitance using Equation 12-10 : [latex]C_{T}= \frac{C_{1}C_{2}}{C_{1}+ C_{2}} [/latex]. Then calculate the total reactance.

[latex]C_{tot}= \frac{C_{1}C_{2}}{C_{1}+ C_{2}} = \frac{(0.01 \ \mu F)(0.068 \ \mu F)}{0.01 \ \mu F + 0.068 \ \mu F}= 0.0087 \ \mu F [/latex]

[latex]X_{C(tot)} = \frac{1}{2\pi fC_{tot}} = \frac{1}{2\pi (5.0 \ kHz)(0.0087 \ \mu F)}= 3.65 \ k\Omega[/latex]

The Parallel Circuit: For the capacitors in parallel in Figure 12-45(b), determine the total reactance from the product-over-sum rule using [latex]X_{C1}[/latex] and [latex] X_{C2}[/latex].

[latex]X_{C(tot)}= \frac{X_{C1}X_{C2}}{X_{C1}+ X_{C2}} = \frac{(3.18 \ k\Omega )(468 \ \Omega )}{3.18 \ k\Omega + 468 \ \Omega} = 408 \ \Omega [/latex]

Question 12.18: What is the total capacitive reactance of each circuit in Fi...

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