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Solution:

a) the resistance re is same with and without source.

b) The mid-band gain without source is Avmid =-72.754 and with source Avs = -54.49

c) Input impedance is the same with and without source.

d) Low cut-off frequency due to source, coupling, and bypass capacitor.

Without source the Low cut-off frequency due to the source capacitor is dominant.

With source the Low cut-off frequency due to bypass capacitor is dominant.

e) Low-frequency plot is shown in the uploaded image.

The Calculation part is shown in the uploaded images.

Solution Given 1 1 (s = 0.47 4F (€ = 2014F Ce=0.47HF Ri=68kn, R2 = 10kh, Rc = 5.6kh, RE= 1.2ks, Re = 3.3kr B=120, Vcc= 14 V Rd) 2 calculation of Lower cut-off frequency Cutoff frequency due to Source Capacitowice (s Ri = RillRailbre 1 fis 21 Rics 204e) Av 3 FLE=85.27 Hz Armial do +22=38.042 fus= 137.76 Midband livel. . -3+ 100 -64 Frequency -388 41 2003 -127 اکرا۔ -18 -21In but impedance (i): No change in input imbedance it is same 2.458 kr d) Lower cut off frequency Source Capacitor is 1 Jis =




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QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

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