---
title: "IC, Q-point, load line, IB = 10 µA"
subject: "Basic Electronics Engineering"
chapter: 3
key: "semester_2/basic_electronics_engineering:ce-ic-q-point-ib-10ua"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/ce-ic-q-point-ib-10ua
site: BCT Engineering
---

# IC, Q-point, load line, IB = 10 µA

Basic Electronics Engineering · Chapter 3: Transistor

- Asked in: 2069 Bhadra, 5 marks
- Video: [Transistor Biasing: What is Q-point? What is Load Line? Fixed Bias Configuration Explained](https://www.youtube.com/watch?v=5T84Jzcgj7M) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/ce-ic-q-point-ib-10ua
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-3

## Question

Find the value of collector current, Q-point and DC load line for common emitter circuit having VCC = 15 V, RC = 10 kΩ, IB = 10 µA and β = 50.

## Short answer

IC = 50 ×10 µA = 0.5 mA, VCE = 15 − 0.5×10k = 10 V. DC load line from (0 V, 1.5 mA) to (15 V, 0 mA). Q-point (10 V, 0.5 mA).

## Answer

**Given:** VCC = 15 V, RC = 10 kΩ, IB = 10 µA, β = 50.

#### Collector current
IC = β IB = 50 × 10 µA = 0.5 mA

#### Q point
VCE = VCC − IC RC = 15 − 0.5 mA × 10 kΩ = 15 − 5 = 10 V

**Q point = (VCE = 10 V, IC = 0.5 mA)**

#### DC load line
From VCC = IC RC + VCE:
- Saturation end (VCE = 0): IC(sat) = 15/10 k = 1.5 mA
- Cut-off end (IC = 0): VCE = 15 V

```text
 IC (mA)
 1.5 |\
     | \
 0.5 |  *Q
     |   \
     +----\------ VCE (V)
         10  15
```

**Answer: IC = 0.5 mA; Q point (10 V, 0.5 mA); the load line joins (0 V, 1.5 mA) and (15 V, 0 mA).**

## Check yourself

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

1. With IB = 10 µA and β = 50, what is IC?
   A. 0.2 mA
   B. 0.5 mA
   C. 5 mA
   D. 1.5 mA
   Answer: B. IC = β IB = 50 × 10 µA = 0.5 mA. (Key point: Collector current)

2. With VCC = 15 V and RC = 10 kΩ, what is VCEQ?
   A. 5 V
   B. 15 V
   C. 10 V
   D. 7.5 V
   Answer: C. VCE = 15 - 0.5 mA × 10k = 10 V. (Key point: Q point)

3. What are the load line end points?
   A. (0 V, 1.5 mA) and (15 V, 0)
   B. (0 V, 0.5 mA) and (10 V, 0)
   C. (0 V, 15 mA) and (10 V, 0)
   D. (15 V, 1.5 mA) and (0, 0)
   Answer: A. IC(sat) = 15/10k = 1.5 mA at VCE = 0 and VCE = VCC = 15 V at IC = 0. (Key point: Load line)

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