ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<
ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<
9th Edition
ISBN: 9781260540666
Author: Hayt
Publisher: MCG CUSTOM
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Textbook Question
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Chapter 9, Problem 1E

For a certain source-free parallel RLC circuit, R = 1 kΩ, C = 3 μF, and L is such that the circuit response is overdamped. (a) Determine a suitable value of L. (b) Write the equation for the voltage v across the resistor if it is known that v(0) = 9 V and dv/dt|t=0+ = 2 V/s.

(a)

Expert Solution
Check Mark
To determine

Find inductor L.

Answer to Problem 1E

The value of inductor L is >12H.

Explanation of Solution

Given data:

Value of resistance R is 1kΩ and

Value of capacitor C is 3μF.

Formula used:

The expression for the two solutions of the characteristic equation of a parallel RLC circuit is as follows:

s1=α+α2ω02 (1)

s2=αα2ω02 (2)

Here,

s1 and s2 is the solutions of the characteristic equation of a parallel RLC circuit.

The value of real part of root of natural response of the source free parallel RLC circuit is:

α=12RC (3)

The value of imaginary part of root of natural response of the source free parallel RLC circuit is:

ω0=1LC (4)

The expression for over damped source free response of parallel RLC circuit is:

α>ω0 (5)

Calculation:

Substitute 1kΩ for R and 3μF for C in equation (3):

α=12×1kΩ×3μF=12×1×103 Ω×3×106F                        { 1μF=106F 1 kΩ=103 Ω}

α=16×103

Substitute 3μF for C in equation (4):

ω0=1L×3μF=1L×3×106F

Substitute 16×103 for α and 1L×3μF for ω0 in equation (5):

16×103>1L×3μF6×103>L×3μF

Squaring both the sides,

36×106>L×3×106FL>36×1063×106FL>12H

Conclusion:

Thus, the value of inductor L is >12H.

(b)

Expert Solution
Check Mark
To determine

Find equation for voltage across resistor.

Answer to Problem 1E

The voltage across resistor is 11.74e120.43t+20.74e212.91tV.

Explanation of Solution

Given data:

Value of initial voltage v(0) is 9V.

Value of rate of change of voltage for t=0+ is 2V/s.

Calculation:

Choose the value of inductor L as 13H.

The general form of natural response for parallel RLC circuit is,

v(t)=A1es1t+A2es2t (6)

Here,

v(t) is the voltage of parallel RLC circuit,

s1 and s2 are the roots of voltage equation and

A1 and A2 are the arbitrary constants.

Differentiate both sides of equation (6) with respect to time:

dv(t)dt=A1s1es1t+A2s2es2t (7)

The expression for the voltage across resistor R in parallel RLC circuit is:

vR(t)=A1es1t+A2es2t (8)

Substitute 0s for t and 9V for v(0) in equation (6):

9V=A1es1×0+A2es2×09V=A1+A2

Rearrange for A1,

A1=9VA2 (9)

Substitute 3μF for C and 13H for L in equation (4):

ω0=113H×3μF=113H×3×106F=16.245×103

ω0=160.128rad/s

Substitute 166.67s1 for α and 160.128rad/s for ω0 in equation (1):

s1=(166.67+(166.67s1)2(160.128rad/s)2) s1=(166.67+27778.8925640.97) s1=(166.67+2137.919) s1=(166.67+46.238) s1

Solve for s1

s1=120.43 s1

Substitute 166.67s1 for α and 160.128rad/s for ω0 in equation (2):

s2=(166.67(166.67s1)2(160.128rad/s)2) s1=(166.6727778.8925640.97) s1=(166.672137.919) s1=(166.6746.238) s1

Solve for s2,

s2=212.91 s1 (10)

Substitute 2V/s for dv(t)dt|t=0, 120.43 for s1 and 212.91 for s2 in equation (7).

2V/s=120.43A1es1×0212.91A2es2×02V/s=120.43A1212.91A2

2V/s=120.43A1+212.91A2 (11)

Substitute 9VA2 for A1 in equation (11):

2V/s=120.43(9VA2)+212.91A22V/s=1083.87V120.43A2+212.91A22V/s=1083.87V+92.48A1

Solve for A1:

2V/s=1083.87V+92.48A1

Rearrange for A1:

92.48A1=1085.87A1=1085.8792.48

A1=11.74

Substitute 11.74 for A1 in equation (9):

11.74=9VA2A2=9V+11.74

A2=20.74

Substitute 120.43 for s1 and 212.91 for s2, 11.74 for A1 and 20.74 for A2 in equation (6).

v(t)=11.74e120.43t+20.74e212.91tV

The voltage across resistor is same as the voltage v(t) due to parallel RLC circuit,

vR(t)=11.74e120.43t+20.74e212.91tV

Conclusion:

Thus, the voltage across resistor is 11.74e120.43t+20.74e212.91tV.

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Chapter 9 Solutions

ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<

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