Chapter 20, Problem 1PP

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

Resistance	Capacitance	Time constant	Total time
150 $k\Omega$	100 $\mu F$
350 $k\Omega$			35 s
	350 pF		10 s
	0.05 $\mu F$
1.2 $M\Omega$	0.47 $\mu F$
	$12\mu F$	0.05 s
86 $k\Omega$			1.5 s
120 $k\Omega$	470 pF
	250 nF		100 ms
	8 $\mu F$		150 $\mu s$
100 $k\Omega$		150 ms
33 $k\Omega$	4 $\mu F$

$\begin{array}{l}\tau =RC\\ R=\frac{\tau }{C}\\ C=\frac{\tau }{R}\end{array}$

$Totaltime=\tau \times 5$

To determine

The missing values in the table.

Answer to Problem 1PP

Resistance	Capacitance	Time Constant	Total Time
150 kΩ	100 µF	15 s	75 s
350 kΩ	20 µF	7s	35 s
142.85 MΩ	350 pF	0.05 s	0.25 s
40 MΩ	0.05 µF	2 s	10 s
1.2 MΩ	0.47 µF	0.564 s	2.82 s
4166.67 Ω	12 µF	0.05 s	0.25 s
86 kΩ	3.488 µF	0.3 s	1.5 s
120 kΩ	470 pF	56.4 µs	282 µs
80 kΩ	250 nF	20 ms	100 ms
3.75 Ω	8 µF	30 µs	150 µs
100 kΩ	1.5 µF	150 ms	750 ms
33 kΩ	4 µF	132 ms	660 ms

Explanation of Solution

We are given the following formulae,

$\begin{array}{l}\tau =RC\mathrm{...}\text{ (1)}\\ R=\frac{\tau }{C}\mathrm{...}\text{ (2)}\\ C=\frac{\tau }{R}\mathrm{...}\text{ (3)}\\ Time=5\tau \mathrm{...}\text{ (4)}\\ \end{array}$

$\begin{array}{l}\left(1\right)\\ \tau =RC=\left(150k\Omega \right)\left(100\mu F\right)=15s\\ Time=5\tau =5\left(15s\right)=75s\\ \\ \left(2\right)\\ \tau =\frac{Time}{5}=\frac{35}{5}=7s\\ C=\frac{\tau }{R}=\frac{\text{ 7}}{350k}=20\mu s\\ \\ \left(3\right)\\ Time=5\tau =5\left(0.05\right)=0.25s\\ R=\frac{\tau }{C}=\frac{\text{ 0}\text{.05}}{350\times {10}^{-12}}=142.85M\Omega \\ \\ \left(4\right)\\ \tau =\frac{Time}{5}=\frac{10}{5}=2s\\ R=\frac{\tau }{C}=\frac{\text{ 2}}{0.05\times {10}^{-6}}=40M\Omega \\ \\ \left(5\right)\\ \tau =RC=\left(1.2M\Omega \right)\left(0.47\mu F\right)=0.564s\\ Time=5\tau =5\left(0.564s\right)=2.82s\\ \\ \left(6\right)\\ Time=5\tau =5\left(0.05\right)=0.25s\\ R=\frac{\tau }{C}=\frac{\text{ 0}\text{.5}}{12\times {10}^{-6}}=4166.67\Omega \end{array}$

$\begin{array}{l}\left(7\right)\\ \tau =\frac{Time}{5}=\frac{1.5}{5}=0.3s\\ C=\frac{\tau }{R}=\frac{\text{ 0}\text{.3}}{86\times {10}^3}=3.488\mu F\\ \\ \left(8\right)\\ \tau =RC=\left(120k\Omega \right)\left(470pF\right)=56.4\mu s\\ Time=5\tau =5\left(56.4\mu s\right)=282\mu s\\ \\ \left(9\right)\\ \tau =\frac{Time}{5}=\frac{100ms}{5}=20ms\\ R=\frac{\tau }{C}=\frac{\text{ 20}\times {\text{10}}^{-3}}{250\times {10}^{-9}}=80k\Omega \\ \\ \left(10\right)\\ \tau =\frac{Time}{5}=\frac{150\mu s}{5}=30\mu s\\ R=\frac{\tau }{C}=\frac{\text{ 30}\times {\text{10}}^{-6}}{8\times {10}^{-6}}=3.75\Omega \\ \\ \left(11\right)\\ Time=5\tau =5\left(150ms\right)=750ms\\ C=\frac{\tau }{R}=\frac{\text{ 150}\times {\text{10}}^{-3}}{100\times {10}^3}=1.5\mu F\\ \\ (12)\\ \tau =RC=\left(33k\Omega \right)\left(4\mu F\right)=132ms\\ Time=5\tau =5\left(15s\right)=660ms\end{array}$