### Figure P6.35The image displays two transistor circuits labeled (c) and (d), each with a different configuration of resistors and voltage sources.#### Circuit (c):- **Transistor Q₃** is present with its base connected to a +1.0 V source.- A 10 kΩ resistor is connected between the +3 V supply and the collector of the transistor.- A 5.6 kΩ resistor is connected between the emitter of the transistor and ground.#### Circuit (d):- **Transistor Q₄** is part of the circuit with its base connected to a +1.5 V source.- An 8.2 kΩ resistor is connected between the +3 V supply and the collector of the transistor.- A 4.7 kΩ resistor is connected between the emitter of the transistor and ground.These circuits demonstrate how different resistor values and input voltages influence the operation of transistor-based circuits. They can be used to explore concepts such as transistor biasing, current flow, and voltage drop across components. **Transcription****Problem 6.35:** For each of the circuits shown in Fig. P6.35, find the emitter, base, and collector voltages and currents. Use β = 50, but assume \( |V_{BE}| = 0.7 \, V \) independent of current level.---**Explanation (for Educational Context):**The problem involves analyzing transistor circuits to find the voltages and currents at the emitter, base, and collector terminals. The parameter β (beta) is the current gain of the transistor, given as 50. The base-emitter voltage (\( V_{BE} \)) is assumed to be a constant 0.7 volts, regardless of the current.Students are expected to apply their knowledge of transistor operation and circuit analysis to determine the specified voltages and currents. This involves understanding the role of each terminal, using Kirchhoff's laws, and applying the given β value to relate collector current to base current.Note: Figure P6.35 referenced in the problem is necessary to provide the complete context and circuit details required for analysis.

[c] Consider the circuit diagram,Given, VB=1VVBE=0.7Vβ=50 Hence, VEB=0.7VE-VB=0.7VE-1=0.7VE=1.7V IE=3-VE10kIE=3-1.710kIE=0.13mA IC=ββ+1IEIC=5050+1×0.13mAIC=0.12745mASince,IE=IC+IB0.13mA=0.12745mA+IBIB=2.549 μA VC=IC×5.6kVC=0.12745mA×5.6kVC=0.71372V[d] Consider the circuit diagram,Given, VB=1.5VVBE=0.7Vβ=50 Hence, VBE=0.7VB-VE=0.71.5-VE=0.7VE=0.8V IE=VE4.7kIE=0.84.7kIE=0.17021mA IC=ββ+1IEIC=5050+1×0.17021mAIC=0.166875 mASince,IE=IC+IB0.17021 mA=0.166875 mA+IBIB=3.335 μA IC=3-VC8.2k0.166875mA=3-VC8.2kVC=3-0.166875mA×8.2kVC=1.631625V

Answered: 6.35 For each of the circuits shown in…

Engineering

Electrical Engineering

6.35 For each of the circuits shown in Fig. P6.35, find the emitter, base, and collector voltages and currents. Use B=50, but assume |VRE = 0.7 V independent of current level.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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