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Materials Science And Engineering
10th Edition
ISBN: 9781119405498
Author: Callister, William D., Jr, RETHWISCH, David G., Jr., 1940- Author.
Publisher: Wiley,
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Chapter 14, Problem 22QAP
To determine
To explain:
The reason for the crystallization of polymer falls with the rise in molecular weight.
Expert Solution & Answer
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Students have asked these similar questions
Design a self-biased JFET circuit (Fig. 6) assuming VGS(0) = -1.3 and ipss=
20 mA. We require a VGS = -0.7. Assume a supply voltage of 15 volts. Draw the
load line for this circuit using Fig. 4b once you have selected the appropriate values
for the components. Does the load line intersect the VGS = -0.7 volt line at the
computed in point?
RD.
RG
Rs
12
20nA
GS = -1.3
VGS
10nA
Fig. 6. Circuit for Examples 2 &3.
50
100
150
200
□ ID(J1)
UDS
Fig. 4b. The IV characteristics of an n-channel JFET (J113). The plots are for VGs increments of
0.05 volts. VGS(0) -1.3. The yellow and blue load lines are for examples 2 &3,
respectively.
Introduction: Orifice and Free Flow Jet in Applied Fluid Mechanics' I need to introduction only for answer
Find the operating point and the load line of a voltage-divider JFET biasing circuit
using the following parameters: VGS(0) = -1.3 and Vcc = 15 volts. Assume
ipss = 20 mA, RG₁ = RG2 = 10 kn, RD = 300, and Rs = 1 kn. Use Fig. 4b for
the IV characteristic of the JFET.
20nA
GS=-1.3 GS
10nA-
50
100
150
200
ID(J1)
UDS
Fig. 4b. The IV characteristics of an n-channel JFET (J113). The plots are for VGs increments of
0.05 volts. VGS(0) -1.3. The yellow and blue load lines are for examples 2 &3,
respectively.
Chapter 14 Solutions
Materials Science And Engineering
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Similar questions
- Design the JFET circuit for the largest in swing. Use the self-bias circuit shown in Fig. 6. Assume that VGS (0) = -1.3 and Vcc = 15 volts. Furthermore, assume that ipss = 20 mA. Using Fig. 4b, draw the load line and identify the Q point. Explain why this will allow the largest swing. Use ip = ipss (1- VGS VGS(0) to show what happens to i, and vps when you have a swing of 0.2 volts in vcs form its operating point (that is, change vas by ±0.2 volts and compute the corresponding iD and VDs). RD RG Rs 0 20nA GS=-1.3 VGS 12 10nA -0- Fig. 6. Circuit for Examples 2 &3. BA-C 50 100 150 200 □ ID(J1) UDS Fig. 4b. The IV characteristics of an n-channel JFET (J113). The plots are for VGs increments of 0.05 volts. VGS(0) -1.3. The yellow and blue load lines are for examples 2 &3, respectively.arrow_forwardplease do the correct VI chrastaristics curve on excel. I am not sure if mine is correctarrow_forwardplease do the correct VI chrastaristics curve on excel. I am not sure if mine is correct. Note the two curves in the picture are for both but its two tries and i dont know which is correct, and probebly both are wrong SCR (Forward Bias Condition) NO VAA VG= 0V, IG=0 mA VG= 5V, IG=4.07mA VG= 10V, IG=9.05mA VAK (V) IAK(mA) VAK (V) IAK(mA) VAK (V) IAK(mA) 1 0 0 0 0 0 0 0 2 5 0.576 4.42 mA 0.576 4.42 mA 0.576 4.43 3 10 7.99 2 0.598 9.4 0.598 9.4 4 15 14.99 0.003 0.612 14.4 0.612 14.4 5 20 19.994 0.004 0.622 19.4 0.622 19.4 6 25 0.63 24.4 0.63 24.4 0.63 24.4 4 30 0.637 29.4 0.637 29.4 0.637 29.4 8 40 0.65 39.4 0.65 39.4 0.65 39.4 9 50 0.66 49.3 0.66 49.3 0.66 49.3 10 60 0.67 59.3 0.67 59.3 0.67 59.3 11 70 0.679 69.3 0.679 69.3 SCR (Reversed Bias…arrow_forward
- compute the load bearing capacity, displacement, and stress distribution, tabulate the answersarrow_forwardcompute the load bearing capacity, displacement, stress distribution, tabulate the answersarrow_forwardBy using the yield line theory, determine the ultimate resisting moment per linear meter (m) for an isotropic reinforced concrete two-way slab to sustain a concentrated factored load of P kN applied as shown in figure. Use equilibrium method in solution Column 2.0 P 8.0 m m m XXXXarrow_forward
- By using the yield line theory, determine the ultimate resisting moment (m) for an isotropic reinforced concrete two-way slab shown in figure under a uniform load (q). Use equilibrium method in solution m m column 20 m 20 marrow_forwardSee both images attachedarrow_forwardQ3: An engine produce 750 kW power and uses gaseous C12H26 as a fuel at 25 C; 200% theoretical air is used and air enters at 500 K. The products of combustion leave at 800 K. The heat loss from the engine is 175 kW. Determine the fuel consumption for complete combustion.arrow_forward
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