MATERIALS SCIENCE+ENGINEERING:WILEY PLUS
10th Edition
ISBN: 9781119815242
Author: Callister
Publisher: WILEY
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Chapter 6, Problem 35QAP
To determine
Ductility in terms of percent reduction in area and in terms of percent elongation.
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Consider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before
the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go-
R
9(1)
i(t)↓
After the switches closes, current flows through the circuit and the capacitor begins to discharge.
The equation that describes the total voltage in the loop comes from Kirchoff's voltage law:
L
di(t)
+ Ri(t)+(0) = 0,
(1)
where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance,
R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of
change of the capacitor charge, and dividing by L, we can write the following homogeneous (no
input source) differential equation for the charge on the capacitor:
4(1) +29(1)+w79(1)=0,
ཀྱི
where
a=
R
2L
and
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9(1) =Aent - Beat,
where
(3)
(4)
(5)
A=
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(82+20)90 +90
and B=
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I need help with this problem and an explanation of the solution for the image described below. (Introduction to Signals and Systems)
Consider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before
the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go.
R
w
i(t)
q(t)
C
н
After the switches closes, current flows through the circuit and the capacitor begins to discharge.
The equation that describes the total voltage in the loop comes from Kirchoff's voltage law:
di(t)
L
+ Ri(t) + (t) = 0,
dt
(1)
where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance,
R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of
change of the capacitor charge, and dividing by L, we can write the following homogeneous (no
input source) differential equation for the charge on the capacitor:
ä(t)+2ag(t)+wg(t) = 0,
(2)
where
R
a
2L
and w₁ = C
LC
The solution to this second order linear differential equation can be written as:
where
81=
q(t) = Ae³¹- Bel
82 =
(3)
(4)
(5)
Chapter 6 Solutions
MATERIALS SCIENCE+ENGINEERING:WILEY PLUS
Ch. 6 - Prob. 1QAPCh. 6 - Prob. 2QAPCh. 6 - Prob. 3QAPCh. 6 - Prob. 4QAPCh. 6 - Prob. 5QAPCh. 6 - Prob. 6QAPCh. 6 - Prob. 7QAPCh. 6 - Prob. 8QAPCh. 6 - Prob. 9QAPCh. 6 - Prob. 10QAP
Ch. 6 - Prob. 11QAPCh. 6 - Prob. 12QAPCh. 6 - Prob. 13QAPCh. 6 - Prob. 14QAPCh. 6 - Prob. 15QAPCh. 6 - Prob. 16QAPCh. 6 - Prob. 17QAPCh. 6 - Prob. 18QAPCh. 6 - Prob. 19QAPCh. 6 - Prob. 20QAPCh. 6 - Prob. 21QAPCh. 6 - Prob. 22QAPCh. 6 - Prob. 23QAPCh. 6 - Prob. 24QAPCh. 6 - Prob. 25QAPCh. 6 - Prob. 26QAPCh. 6 - Prob. 27QAPCh. 6 - Prob. 28QAPCh. 6 - Prob. 29QAPCh. 6 - Prob. 30QAPCh. 6 - Prob. 31QAPCh. 6 - Prob. 32QAPCh. 6 - Prob. 33QAPCh. 6 - Prob. 34QAPCh. 6 - Prob. 35QAPCh. 6 - Prob. 36QAPCh. 6 - Prob. 37QAPCh. 6 - Prob. 38QAPCh. 6 - Prob. 39QAPCh. 6 - Prob. 40QAPCh. 6 - Prob. 41QAPCh. 6 - Prob. 42QAPCh. 6 - Prob. 43QAPCh. 6 - Prob. 44QAPCh. 6 - Prob. 45QAPCh. 6 - Prob. 46QAPCh. 6 - Prob. 47QAPCh. 6 - Prob. 48QAPCh. 6 - Prob. 49QAPCh. 6 - Prob. 50QAPCh. 6 - Prob. 51QAPCh. 6 - Prob. 52QAPCh. 6 - Prob. 53QAPCh. 6 - Prob. 54QAPCh. 6 - Prob. 55QAPCh. 6 - Prob. 56QAPCh. 6 - Prob. 57QAPCh. 6 - Prob. 58QAPCh. 6 - Prob. 59QAPCh. 6 - Prob. 1DPCh. 6 - Prob. 2DPCh. 6 - Prob. 3DPCh. 6 - Prob. 4DPCh. 6 - Prob. 1SSPCh. 6 - Prob. 1FEQPCh. 6 - Prob. 2FEQPCh. 6 - Prob. 3FEQPCh. 6 - Prob. 4FEQPCh. 6 - Prob. 5FEQP
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