Chapter 21, Problem 1PP

Capacitive Circuits

Fill in all the missing values. Refer to the formulas that follow.

$\begin{array}{l}{X}_C=\frac{1}{2\pi fC}\\ C=\frac{1}{2\pi f{X}_C}\\ f=\frac{1}{2\pi C{X}_c}\end{array}$

Capacitance	$X_C$	Frequency
38 $\mu F$		60 Hz
	78.8 $\Omega$	400 Hz
250 pF	4.5 $k\Omega$
234 $\mu F$		10 kHZ
	240 $\Omega$	50 Hz
10 $\mu F$	36.8 $\Omega$
560 nF		2 MHz
	15 $k\Omega$	60 Hz
75 nF	560 $\Omega$
470 pF		200 kHz
	6.8 $k\Omega$	400 Hz
34 $\mu F$	450 $\Omega$

To determine

The missing values.

Answer to Problem 1PP

Capacitance	Xc	Frequency
38 µF	69.8 Ω	60 Hz
5.05 µF	78.8 Ω	400 Hz
250 pF	4.5 kΩ	141.47 kHz
234 µF	0.068 Ω	10 kHz
13.26 µF	240 Ω	50 Hz
10 µF	36.8 Ω	432.48 Hz
560 nF	0.1421 Ω	2 MHz
176.8 nF	15 kΩ	60 Hz
75 µF	560 Ω	3.78 Hz
470 pF	1.69 kΩ	200KHz
58 nF	6.8 kΩ	400Hz
34 µF	450 Ω	10.4 Hz

Explanation of Solution

Description:

We are given the following formulae,

$\begin{array}{l}{X}_c=\frac{1}{\text{2}\pi fC}\mathrm{...}\text{(1)}\\ C=\frac{1}{\text{2}\pi f{X}_c}\mathrm{...}\text{(2)}\\ f=\frac{1}{\text{2}\pi C{X}_c}\mathrm{...}\text{(3)}\\ \end{array}$

$\begin{array}{l}\left(1\right)X_c=\frac{1}{\text{2}\pi fC}=\frac{1}{\text{2}\pi \times 60\times 38\times {10}^{-6}}=69.80\Omega \\ \left(2\right)C=\frac{1}{\text{2}\pi f{X}_c}=\frac{1}{\text{2}\pi \times 400\times 78.8}=5.05\mu F\\ \left(3\right)f=\frac{1}{\text{2}\pi C{X}_c}=\frac{1}{\text{2}\pi \times 250\times {10}^{-12}\times 4.5\times {10}^3}=141.47kHz\\ \left(4\right)X_c=\frac{1}{\text{2}\pi fC}=\frac{1}{\text{2}\pi \times 10\times {10}^3\times 234\times {10}^{-6}}=0.068\Omega \\ \left(5\right)C=\frac{1}{\text{2}\pi f{X}_c}=\frac{1}{\text{2}\pi \times 50\times 240}=13.26\mu F\\ \left(6\right)f=\frac{1}{\text{2}\pi C{X}_c}=\frac{1}{\text{2}\pi \times 10\times {10}^{-6}\times 36.8}=432.48Hz\\ \left(7\right)X_c=\frac{1}{\text{2}\pi fC}=\frac{1}{\text{2}\pi \times 560\times {10}^{-9}\times 2\times {10}^6}=0.1421\Omega \\ \left(8\right)C=\frac{1}{\text{2}\pi f{X}_c}=\frac{1}{\text{2}\pi \times 60\times 15\times {10}^3}=176.8nF\\ \left(9\right)f=\frac{1}{\text{2}\pi C{X}_c}=\frac{1}{\text{2}\pi \times 75\times {10}^{-6}\times 560}=3.78Hz\\ \left(10\right)X_c=\frac{1}{\text{2}\pi fC}=\frac{1}{\text{2}\pi \times 200\times {10}^3\times 470\times {10}^{-12}}=1.69k\Omega \\ \left(11\right)C=\frac{1}{\text{2}\pi f{X}_c}=\frac{1}{\text{2}\pi \times 400\times 6.8\times {10}^3}=58nF\\ \left(12\right)f=\frac{1}{\text{2}\pi C{X}_c}=\frac{1}{\text{2}\pi \times 34\times {10}^{-6}\times 450}=10.4Hz\end{array}$