Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN: 9781337900348
Author: Stephen L. Herman
Publisher: Cengage Learning
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Chapter 19, Problem 1PP

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

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

τ = R C R = τ C C = τ R

T o t a l t i m e = τ × 5

Expert Solution & Answer
Check Mark
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,

τ=RC       ... (1)R=τC        ... (2)C=τR        ... (3)Time=5τ   ... (4) 

(1) τ=RC=(150kΩ)(100μF)=15sTime=5τ=5(15s)=75s(2) τ=Time5=355=7s     C= τR= 7350k=20μs(3) Time=5τ=5(0.05)=0.25s R= τC= 0.05350×1012=142.85MΩ(4) τ=Time5=105=2s     R= τC= 20.05×106=40MΩ(5) τ=RC=(1.2MΩ)(0.47μF)=0.564sTime=5τ=5(0.564s)=2.82s(6) Time=5τ=5(0.05)=0.25s R= τC= 0.512×106=4166.67Ω

(7)τ=Time5=1.55=0.3s     C= τR= 0.386×103=3.488μF(8)  τ=RC=(120kΩ)(470pF)=56.4μsTime=5τ=5(56.4μs)=282μs(9) τ=Time5=100ms5=20ms     R= τC= 20×103250×109=80kΩ(10) τ=Time5=150μs5=30μs     R= τC= 30×1068×106=3.75Ω(11) Time=5τ=5(150ms)=750msC= τR= 150×103100×103=1.5μF(12)τ=RC=(33kΩ)(4μF)=132msTime=5τ=5(15s)=660ms

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