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Materials Science and Engineering
9th Edition
ISBN: 9781118324578
Author: Jr. William D. Callister
Publisher: WILEY
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Chapter 16.16, Problem 23QP
To determine
To State:
The distinction between carbon and graphite.
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HW_03.pdf EE 213-01 > Assignments
HW_#3
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b Success Confirmation of Question Submission | bartleby
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1) (5 pts) Use node voltage analysis at nodes A and B to determine values for VA, VB, I₁, 12 and 13.
If a current has a negative value, then the reference direction chosen was incorrect and the current
is positive in the opposite direction. Your answer can be the negative value – i.e. you find that 12 =
-1mA. Note, the current through R5 is (VB - 28 Volts)/10K
10K VA
20K VB
10K
R1
12
R2
R5
13
IS1
R3
10 mA
10K
28 V
R4
VS1
5K
-VREF
B
2) (5pts) Use node voltage analysis to find values for the node voltages VA, VR and Vc. Use these
node voltages to determine values for V1, V2, V3, V4, 11, 12, 13, and 14. Hints: V3 = VB and the value
for Vc can be found with no analysis. Dependent current source has a value of gV3 amps
VA
gV3 Amps
14
12
20K
+
V2
-
+
10K
VB
13
V3 10K
-
20K
I₁
V₁ +
Vc
g=5x10-5
20 volts
VREF
3) (5 pts) For the…
5. Problem 8-46 on Page 374.
8-46. The L-shaped frame is made from two fixed-connected
segments. Determine the vertical displacement of the end C.
Use the method of virtual work. El is constant.
-9 ft-
2 k/ft
12 ft
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1) (5 pts)For the circuit shown, find the value of Ia, Ib, I¸ and Va:
Ib
10 Ohms
+
Ic
20 Ohms
70 Volts
Va
40 Ohms
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Hint: use KCL to find Ic in terms of la and lb, use KVL around the right loop to find a relationship
between la and lb, use KVL around the outer loop to solve for a current value (use ohms law for
the voltage drops across the resistors)
-
2) (5 pts) For problem 1, show that the power supplied (delivered) by the source equals the
power absorbed by the resistors.
3) (5 pts) Determine the equivalent resistance looking into terminals a-b for the two circuits
shown. Use series and parallel resistor combinations. Hint: work from the opposite end of
terminals a-b towards terminals a-b
a)
24 Ohms
10 Ohms
8 Ohms
a
REQ
b)
b =
75K
5 Ohms
50K
60 Ohms
4 OHMs
20 Ohms
30 Ohms
□ a
20K
8.2K
40K
16K
10K
□ b
7.2K
REQ
Chapter 16 Solutions
Materials Science and Engineering
Ch. 16.16 - Prob. 1QPCh. 16.16 - Prob. 2QPCh. 16.16 - Prob. 3QPCh. 16.16 - Prob. 4QPCh. 16.16 - Prob. 5QPCh. 16.16 - Prob. 6QPCh. 16.16 - Prob. 7QPCh. 16.16 - Prob. 8QPCh. 16.16 - Prob. 9QPCh. 16.16 - Prob. 10QP
Ch. 16.16 - Prob. 11QPCh. 16.16 - Prob. 12QPCh. 16.16 - Prob. 13QPCh. 16.16 - Prob. 14QPCh. 16.16 - Prob. 15QPCh. 16.16 - Prob. 16QPCh. 16.16 - Prob. 17QPCh. 16.16 - Prob. 18QPCh. 16.16 - Prob. 19QPCh. 16.16 - Prob. 20QPCh. 16.16 - Prob. 21QPCh. 16.16 - Prob. 22QPCh. 16.16 - Prob. 23QPCh. 16.16 - Prob. 24QPCh. 16.16 - Prob. 25QPCh. 16.16 - Prob. 26QPCh. 16.16 - Prob. 27QPCh. 16.16 - Prob. 28QPCh. 16.16 - Prob. 29QPCh. 16.16 - Prob. 30QPCh. 16.16 - Prob. 2DPCh. 16.16 - Prob. 3DPCh. 16.16 - Prob. 4DPCh. 16.16 - Prob. 5DPCh. 16.16 - Prob. 6DPCh. 16.16 - Prob. 1FEQPCh. 16.16 - Prob. 2FEQPCh. 16.16 - Prob. 3FEQPCh. 16.16 - Prob. 4FEQP
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