Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Mechanical Engineering homework question answer, step 1, image 1 Mechanical Engineering homework question answer, step 1, image 2 Step 2 Mechanical Engineering homework question answer, step 2, image 1 Step 3 Mechanical Engineering homework question answer, step 3, image 1

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Solution:

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

b) The mid-band gain without source is A_vmid =-72.754 and with source A_vs = -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 Rs = 0.82 km2, VEE=0.7V Calculation without Rs and signal source Vs a) To determine re for Dc Conditions Using approximate analysis BRE > 10R2 120 x 162X.103 > 10 x 10 x103 144 x103 > 100x10 The result is 10x10² x 14 Ball voltage VB = Rg Vcc Ritra = 179+V Tox 103 +68x10? The Emitter Current VE IE RE 1794-0.7 0.911mA 10 2x103 26x103 3 26 28:54 0.911xio The resistance re *** IE Midband gain Vo Armid = Roller re -5.6x10|| 3-3x10 72.754 28.540 o The input impedance (27) Z1 = Rp a Rillrall Bre Bre 120 X 26.4 = Zi - 68x103 || 10x103 || 3-42uxiol 2.458 km

d) 2 calculation of Lower cut-off frequency Cutoff frequency due to Source Capacitowice (s Ri = RillRailbre 1 fis 21 Rics 204 58kr. 1 & X 2.458X103 x 0.47x166 Cs fus = 137. 76 Hz 1 Vi & Rille, Bre J Fig: Equivalent circuit for is Cut-off friquenly due to coupling capacitare se. (c) Cc fie T 5 T ㅠ }R, vo 2IT(R(TRL) CC Rc YO . dir (s.6x10°+3.3 *10%)x0-47xjós 38.048 HZ Fig: Equivalent circuit for & 2 Cut off frequency due to Bypass Capacitana (Rille2]+ 1 FLE ZIT RECE 1.263 RE CEOZOKF Re = RE tre B (68k 11 lok - 1.2x10 t 28.54 120 Re= 93.3132 RillR2 1110k + 28.54) 2x 103 111 1166** 1 6 11 FLE & IT Re CE 211 x 93.313 x 20 x 10 = 85.274 HZ

e) 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 -21 -247 -21 - Fig: Low frequency plot for the given Circuit Calculation of low frequency - 38.04H2 FLE= = 85.27H2 fis = 137.76H2 Low Cut-off frequency is dominant due to fus = 137.76H2. Calculation with Rs and signal source Vs fuc 1 is same @ The resistance re = 28.542 b) Midband gain Vo - 72,754 Armid = vi Rs The input voltage. vi Vis Rivs Ri Ritrs y 2-458x103 Vi Vs - 0.7498 2.458X103 +0.82x107 fig: With Rs fus The Gain with source vs Vo Ars = Vo Vi - 6- 72-754)x0.7498 VS 2-54.49 Gain Reduces Armid Avs

In but impedance (i): No change in input imbedance it is same 2.458 kr d) Lower cut off frequency Source Capacitor is 1 Jis = La Ceit Rs.cs 27(2-456X163 +0+62 813)X0-47x16 tus 103.303 H2 Coupling Capacitor fuc LIT CRc+R() (c 38.048 HZ Lower cut off frequency due to cc. No change in Bypass capacitor 1 FLE = & IT Rece Where Re = 1 Rs ß ( 1449.5 +28 54 Rs = Rsl Rillea - 0.82x10? 11 68x18? ! cxid? - 810.229 #oxic) Re = 1.2k G 28:54 120 - sto 2k11 34.785 749.52 33.gos - 235.401 H2 } FLE } - 6 21 X 33.8 X 20x10 x 2T ReCE e The Low-Cut off frequency fus = 103.303 42, fuc= 35.048H2 FLE=235.401 42 The low cut off frequenly is dominant due to fLE = 235-401 H2