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(b) Repeat part (a) if the 10-kN resistance is changed to 30 kN. (c) Repeat part (a) if the 10-kN resistance is changed to 60 kN. Hint: apply Thévenin's theorem to the net- work seen by D,'s anode. (d) Repeat part (a) if the 15-kN resistance is re- moved from the circuit

(b) Repeat part (a) if the 10-kN resistance is changed to 30 kN. (c) Repeat part (a) if the 10-kN resistance is changed to 60 kN. Hint: apply Thévenin's theorem to the net- work seen by D,'s anode. (d) Repeat part (a) if the 15-kN resistance is re- moved from the circuit

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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(b) Repeat part (a) if the 10-kN resistance is changed to 30 kN. (c) Repeat part (a) if the 10-k2 resistance is changed to 60 kN. Hint: apply Thévenin's theorem to the net- work seen by D,'s anode. (d) Repeat part (a) if the 15-kN resistance is re- moved from the circuit.
Transcribed Image Text:(b) Repeat part (a) if the 10-kN resistance is changed to 30 kN. (c) Repeat part (a) if the 10-k2 resistance is changed to 60 kN. Hint: apply Thévenin's theorem to the net- work seen by D,'s anode. (d) Repeat part (a) if the 15-kN resistance is re- moved from the circuit.
1.5 (a) Find the current and voltage of each diode in the circuit of Fig. P1.5. 15 V 30 kΩ 10 kΩ D1 D2 15 kN 20 kΩ D3 V -15 V FIGURE P1.5
Transcribed Image Text:1.5 (a) Find the current and voltage of each diode in the circuit of Fig. P1.5. 15 V 30 kΩ 10 kΩ D1 D2 15 kN 20 kΩ D3 V -15 V FIGURE P1.5
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Let us find the find the voltages seen by the diode terminale 15 V 15-30 i Isi=-15 VB = -15V 30 k12 10 k12 U 45 B В 30 v=15vso, Lets assume all diodes are on so VA = VB = uc in this case and =0, so, VA = U₂ =ų to Now, marking currents,15 V 15-0 = = 0.5mA 15-0 =1.5mA 30k lok 30 k 12 10 k 12 Joos OV D2 D # 41-5mA OV ov 15 k12 20 k 12 M D3 V 0-(-15)_ l mA 10-1-Diode D, has to How 0.5mA in reverse direction to satisfy this . so, our assumption of Diode D, ON is invalid , so So Da D, o15 V 9 15-0 = 2 15-(-15) 0.67MA i 10K (15+30k 30 k12 10 k12 1.5mA Yl-SMA 30 VA D N* Ou D2 KOU = i= = 0.67mA A ON 100751 45k yCurrent through De is 0.75mA current through I is 0.75MA D, off so current =O Ao 15 V Assume all Diodes are on YO5mA 4o-5MA 30kh tom 30 k 12 Us VA D TA D2 * Uc VO:75mA IMA 15 k12 20 k 12 D3 V = -15 VD, and D₂ need to flow current in reverse direction so Assumption for them is invalid so D, — OFF D • off De JON 315 V KUL 15-301-201=-15 30k ΑΝΩ 30 k 12 soi=30 D2 D *Atx off i i = 0.6mA ON Dz current 15 k12 20 k12 } D3 Voff -15 V15 V Same as previous part D, and I are off Gok Η ΕΩ 2 30 kΩ D2 D A * 0 15-(-15) 60 k+ 206 not 15 k12 20 k 12 } D3 V - 0.375m15 V let all Diodes are on Yo 10-5mA 1.5mA 30 k12 10 k12 0.5D, D2 0-15 ou OV OU KA Renouah Po 75mA lc = 1.25mA D3 V 154 20 k

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