Page 1 of 184
IBHRE CEPS EXAM PREP 2026-2027:
300+ PRACTICE QUESTIONS WITH
VERIFIED ANSWERS & RATIONALES |
CARDIAC ELECTROPHYSIOLOGY
SPECIALIST CERTIFICATION | A+
SOLUTIONS GUARANTEED PASS
# TABLE OF CONTENTS
| Section | Topic | Weight | Questions |
|---------|-------|--------|-----------|
| **I** | **Physics of Electrophysiology** | 5% | 15 |
| **II** | **Cardiac Anatomy and Physiology** | 10% | 28 |
| **III** | **Pharmacology** | 6% | 17 |
| **IV** | **Fundamentals of Electrophysiology** | 20% | 52 |
| **V** | **Clinical Assessment** | 8% | 22 |
| **VI** | **Laboratory Considerations** | 6% | 16 |
| **VII** | **Invasive Electrophysiology** | 30% | 75 |
| **VIII** | **Implantable Devices** | 5% | 15 |
| **IX** | **Real-time and Diagnostic Imaging** | 5% | 14 |
| **X** | **Research Methodology & Interpretation** | 1% | 4 |
| **XI** | **Comprehensive Review Questions** | — | 42 |
,Page 2 of 184
# SECTION I: PHYSICS OF ELECTROPHYSIOLOGY (5%)
## Question 1
A 12-lead ECG machine is set to a paper speed of 25 mm/s and a gain of 10 mm/mV. A QRS
complex measures 2 small squares in duration and 10 small squares in amplitude. What is the
correct calculation of the heart rate if the R-R interval measures 20 small squares?
A. 60 beats per minute
B. 75 beats per minute
C. 100 beats per minute
D. 150 beats per minute
**Correct Answer: B**
**Rationale:** At 25 mm/s paper speed, each small square (1 mm) represents 0.04 seconds, and
each large square (5 mm) represents 0.20 seconds. The R-R interval of 20 small squares equals
20 × 0.04 = 0.80 seconds. Heart rate = 60/0.80 = 75 beats per minute. The QRS duration (2 small
squares = 0.08 seconds) and amplitude (10 small squares = 10 mm = 1 mV at standard gain) are
distractors for this calculation.
---
## Question 2
According to Ohm's law, if a pacing system delivers a current of 2 mA through a myocardial
tissue resistance of 500 Ω, what is the voltage delivered?
A. 0.004 V
B. 0.25 V
C. 1.0 V
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D. 250 V
**Correct Answer: C**
**Rationale:** Ohm's law states V = I × R. With I = 2 mA = 0.002 A and R = 500 Ω, V = 0.002
× 500 = 1.0 V. Understanding Ohm's law is fundamental to electrophysiology as it governs the
relationship between voltage, current, and resistance in all cardiac electronic devices.
---
## Question 3
Which filter setting would be most appropriate for recording far-field ventricular electrograms
while minimizing low-frequency artifact from respiration?
A. High-pass filter at 0.05 Hz
B. High-pass filter at 30 Hz
C. Low-pass filter at 500 Hz
D. Band-pass filter 0.5–100 Hz
**Correct Answer: B**
**Rationale:** A high-pass filter at 30 Hz effectively attenuates low-frequency signals (<30 Hz)
such as respiration artifact (typically 0.1–1 Hz) while preserving the higher-frequency
components of ventricular electrograms. Option A (0.05 Hz) would allow respiration artifact to
pass. Option C describes a low-pass filter which removes high frequencies. Option D's range
(0.5–100 Hz) would still allow some low-frequency noise.
---
, Page 4 of 184
## Question 4
The biophysics of RF ablation involves resistive heating of tissue. Which of the following
statements correctly describes the relationship between power delivery and lesion size?
A. Lesion size increases linearly with power regardless of impedance
B. Lesion size is inversely proportional to the square of the current
C. Lesion size is directly proportional to the power delivered and duration, but limited by
impedance rise
D. Lesion size is independent of electrode-tissue contact
**Correct Answer: C**
**Rationale:** RF ablation lesion size is determined by the power (P = I² × R) delivered and the
duration of application, but is limited by impedance rise which occurs when tissue desiccates and
forms coagulum. Lesion size does not increase linearly with power due to these limiting factors.
Electrode-tissue contact is critical for effective energy delivery.
---
## Question 5
During an electrophysiology study, a bipolar electrogram is recorded with a peak-to-peak
amplitude of 4 mV. If the gain is increased from 10 mm/mV to 20 mm/mV, what will be the new
displayed amplitude?
A. 2 mm
B. 4 mm
C. 40 mm
D. 80 mm
IBHRE CEPS EXAM PREP 2026-2027:
300+ PRACTICE QUESTIONS WITH
VERIFIED ANSWERS & RATIONALES |
CARDIAC ELECTROPHYSIOLOGY
SPECIALIST CERTIFICATION | A+
SOLUTIONS GUARANTEED PASS
# TABLE OF CONTENTS
| Section | Topic | Weight | Questions |
|---------|-------|--------|-----------|
| **I** | **Physics of Electrophysiology** | 5% | 15 |
| **II** | **Cardiac Anatomy and Physiology** | 10% | 28 |
| **III** | **Pharmacology** | 6% | 17 |
| **IV** | **Fundamentals of Electrophysiology** | 20% | 52 |
| **V** | **Clinical Assessment** | 8% | 22 |
| **VI** | **Laboratory Considerations** | 6% | 16 |
| **VII** | **Invasive Electrophysiology** | 30% | 75 |
| **VIII** | **Implantable Devices** | 5% | 15 |
| **IX** | **Real-time and Diagnostic Imaging** | 5% | 14 |
| **X** | **Research Methodology & Interpretation** | 1% | 4 |
| **XI** | **Comprehensive Review Questions** | — | 42 |
,Page 2 of 184
# SECTION I: PHYSICS OF ELECTROPHYSIOLOGY (5%)
## Question 1
A 12-lead ECG machine is set to a paper speed of 25 mm/s and a gain of 10 mm/mV. A QRS
complex measures 2 small squares in duration and 10 small squares in amplitude. What is the
correct calculation of the heart rate if the R-R interval measures 20 small squares?
A. 60 beats per minute
B. 75 beats per minute
C. 100 beats per minute
D. 150 beats per minute
**Correct Answer: B**
**Rationale:** At 25 mm/s paper speed, each small square (1 mm) represents 0.04 seconds, and
each large square (5 mm) represents 0.20 seconds. The R-R interval of 20 small squares equals
20 × 0.04 = 0.80 seconds. Heart rate = 60/0.80 = 75 beats per minute. The QRS duration (2 small
squares = 0.08 seconds) and amplitude (10 small squares = 10 mm = 1 mV at standard gain) are
distractors for this calculation.
---
## Question 2
According to Ohm's law, if a pacing system delivers a current of 2 mA through a myocardial
tissue resistance of 500 Ω, what is the voltage delivered?
A. 0.004 V
B. 0.25 V
C. 1.0 V
,Page 3 of 184
D. 250 V
**Correct Answer: C**
**Rationale:** Ohm's law states V = I × R. With I = 2 mA = 0.002 A and R = 500 Ω, V = 0.002
× 500 = 1.0 V. Understanding Ohm's law is fundamental to electrophysiology as it governs the
relationship between voltage, current, and resistance in all cardiac electronic devices.
---
## Question 3
Which filter setting would be most appropriate for recording far-field ventricular electrograms
while minimizing low-frequency artifact from respiration?
A. High-pass filter at 0.05 Hz
B. High-pass filter at 30 Hz
C. Low-pass filter at 500 Hz
D. Band-pass filter 0.5–100 Hz
**Correct Answer: B**
**Rationale:** A high-pass filter at 30 Hz effectively attenuates low-frequency signals (<30 Hz)
such as respiration artifact (typically 0.1–1 Hz) while preserving the higher-frequency
components of ventricular electrograms. Option A (0.05 Hz) would allow respiration artifact to
pass. Option C describes a low-pass filter which removes high frequencies. Option D's range
(0.5–100 Hz) would still allow some low-frequency noise.
---
, Page 4 of 184
## Question 4
The biophysics of RF ablation involves resistive heating of tissue. Which of the following
statements correctly describes the relationship between power delivery and lesion size?
A. Lesion size increases linearly with power regardless of impedance
B. Lesion size is inversely proportional to the square of the current
C. Lesion size is directly proportional to the power delivered and duration, but limited by
impedance rise
D. Lesion size is independent of electrode-tissue contact
**Correct Answer: C**
**Rationale:** RF ablation lesion size is determined by the power (P = I² × R) delivered and the
duration of application, but is limited by impedance rise which occurs when tissue desiccates and
forms coagulum. Lesion size does not increase linearly with power due to these limiting factors.
Electrode-tissue contact is critical for effective energy delivery.
---
## Question 5
During an electrophysiology study, a bipolar electrogram is recorded with a peak-to-peak
amplitude of 4 mV. If the gain is increased from 10 mm/mV to 20 mm/mV, what will be the new
displayed amplitude?
A. 2 mm
B. 4 mm
C. 40 mm
D. 80 mm
