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Name: ____________________ Date: ____________
1.Easy
This lead II rhythm is regular. Count the small boxes between two consecutive R waves and use the 1500 method to find the heart rate in beats per minute, rounded to a whole number.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
heart rate = bpm
2.Easy
Measure the PR interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
PR = s
3.Medium
Measure the QT interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QT = s
4.Medium
Interpret this lead II rhythm strip. Which rhythm is it? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1atrial flutter, 4:1 conduction
2sinus rhythm with PVCs
3junctional rhythm
4atrial fibrillation
5.Hard
This lead II strip shows a conduction problem between the atria and the ventricles. Which rhythm is it? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1third-degree (complete) AV block
2first-degree AV block
3second-degree AV block type II
4junctional rhythm
6.Hard
A monitored patient is found unresponsive and not breathing normally, and no carotid pulse can be felt. The monitor shows this organised rhythm. What is this situation called? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1sinus bradycardia
2ventricular fibrillation
3pulseless electrical activity (PEA)
4asystole
Answer key
1. heart rate = 48 bpm
2. PR = 0.2 s
3. QT = 0.4 s
4. atrial fibrillation
5. third-degree (complete) AV block
6. pulseless electrical activity (PEA)
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