---
title: "BJT ICQ and VCEQ with RB 100k, RC 3k"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:bjt-icq-vceq-rb100k"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bjt-icq-vceq-rb100k
site: BCT Engineering
---

# BJT ICQ and VCEQ with RB 100k, RC 3k

Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor

- Asked in: 2081 Baisakh, 4 marks
- Video: [[Solved]- Numerical for a BJT (Ib, Ic, Vce & Vcb =? )](https://www.youtube.com/watch?v=gu_FYvL5zc4) — Electronics Circuit Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bjt-icq-vceq-rb100k
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Determine the dc collector current ICQ and collector emitter voltage VCEQ for the following circuit. Given β = 100.
[Figure: NPN transistor with base resistor RB = 100 kΩ from source VBB = 3 V to base; collector resistor RC = 3 kΩ from a 10 V supply to collector; emitter grounded]

## Short answer

IB = (3-0.7)/100kΩ = 23μA. ICQ = β IB = 2.3mA. VCEQ = VCC - ICQ RC = 10 - 6.9 = 3.1V. The transistor is in active region.

## Answer

Use the base-emitter loop to find IB, then IC = βIB, then the collector loop for VCE. Assume the active region (to be checked) and VBE = 0.7 V.

#### Given
VBB = 3 V, RB = 100 kΩ, RC = 3 kΩ, VCC = 10 V, β = 100.

#### Base loop
```text
IB = (VBB - VBE)/RB = (3 - 0.7)/100 kΩ = 23 µA
```

#### Collector current
```text
ICQ = β IB = 100 × 23 µA = 2.3 mA
```

#### Collector-emitter voltage
```text
VCEQ = VCC - ICQ RC = 10 - (2.3 mA)(3 kΩ) = 10 - 6.9 = 3.1 V
```

#### Check
VCEQ = 3.1 V > VCE(sat) ≈ 0.2 V, so the transistor is in the active region and the assumption is valid.

**Answer: ICQ = 2.3 mA, VCEQ = 3.1 V.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is IB for VBB = 3 V, RB = 100 kΩ, VBE = 0.7 V?
   A. 7 µA
   B. 23 µA
   C. 30 µA
   D. 2.3 mA
   Answer: B. IB = (3 - 0.7)/100 kΩ = 23 µA. (Key point: Base current)

2. What is ICQ with beta = 100?
   A. 23 µA
   B. 0.23 mA
   C. 3 mA
   D. 2.3 mA
   Answer: D. ICQ = beta IB = 100 × 23 µA = 2.3 mA. (Key point: Collector current)

3. What is VCEQ with RC = 3 kΩ and VCC = 10 V?
   A. 10 V
   B. 3.1 V
   C. 6.9 V
   D. 0.2 V
   Answer: B. VCEQ = 10 - 2.3 mA × 3 kΩ = 3.1 V. (Key point: VCEQ)

4. How is the assumed active region verified?
   A. VCEQ (3.1 V) is greater than VCE(sat) ≈ 0.2 V
   B. IB is greater than IC
   C. VCEQ equals VCC
   D. VBE is greater than VCE
   Answer: A. Since VCE exceeds the saturation value, the active-region assumption is valid. (Key point: Region check)

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