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(a) The voltage divider biasing of BJT is shown in Fig. Q2 (a). Here, R1=39 KQ, R2 = 3.9 KQ, Rc= 10 KQ, RE = 1.5 KN, B= (125+X) and Vcc= (X+15) V. Where X=Last digit of your ID+2. %3D Calculate i. ii. Ig and Ic VCE and VBC VB and VE iii. + Vcco R1 Rc B E R2 RE Fig.Q2(a) ww ww

(a) The voltage divider biasing of BJT is shown in Fig. Q2 (a). Here, R1=39 KQ, R2 = 3.9 KQ, Rc= 10 KQ, RE = 1.5 KN, B= (125+X) and Vcc= (X+15) V. Where X=Last digit of your ID+2. %3D Calculate i. ii. Ig and Ic VCE and VBC VB and VE iii. + Vcco R1 Rc B E R2 RE Fig.Q2(a) ww ww

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Last digit of my id = 2, so X=2

2. (a) The voltage divider biasing of BJT is shown in Fig. Q2 (a). Here, R1=39 KN, R2 = 3.9 KN, Rc= 10 KQ, RE = 1.5 KN, B= (125+X) and Vcc= (X+15) V. Where X=Last digit of your ID+2. Calculate Ig and Ic VCE and VBC VB and VE i. ii. 111. + Vc R1 Rc R2 RE Fig.Q2(a) ww- ww ww
Transcribed Image Text:2. (a) The voltage divider biasing of BJT is shown in Fig. Q2 (a). Here, R1=39 KN, R2 = 3.9 KN, Rc= 10 KQ, RE = 1.5 KN, B= (125+X) and Vcc= (X+15) V. Where X=Last digit of your ID+2. Calculate Ig and Ic VCE and VBC VB and VE i. ii. 111. + Vc R1 Rc R2 RE Fig.Q2(a) ww- ww ww
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