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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Question:

None

Explanation

Given Information:

  VCC=3.3 V, RE=500 Ω , RC=4 kΩ, R1=85 kΩ, R2=35 kΩ and β=150$V_{CC}=3.3\text{ V, }{R}_E\text{=500 }\Omega \text{ , }{R}_C\text{=4 k}\Omega ,R_1\text{=85 k}\Omega ,R_2\text{=35 k}\Omega \text{ and }\beta =150$

Calculation:

Use the equations of Thevenin voltage and Thevenin resistance to solve the problem. The figure below shows the original circuit and the

  

Calculation:

Thevenin resistance is

  RTH=(R1R2R1+R2)=(85×3585+35) kΩ$R_{TH}=\left(\frac{R_1{R}_2}{R_1+R_2}\right)=\left(\frac{85\times 35}{85+35}\right)\text{ k}\Omega$

  RTH=24.8 kΩ$R_{TH}=24.8\text{ k}\Omega$

Thevenin voltage is

  VTH=(R2R1+R2)VCC=(3585+35)3.3 V$V_{TH}=\left(\frac{R_2}{R_1+R_2}\right)V_{CC}=\left(\frac{35}{85+35}\right)3.3\text{ V}$

  VTH=0.9625 V