ECG Basics and Lead Placement Practice Questions

32 multiple-choice questions on ECG basics and lead placement for paramedic and EMT students, from Chapter 18 (Cardiovascular Emergencies). Read each question, pick your answer, then tap Show answer to check it and read the explanation.

32 practice questions

1. At the standard paper speed of 25 mm/sec, each small box on the ECG represents:

   1. A. 0.04 seconds
   2. B. 0.02 seconds
   3. C. 0.20 seconds
   4. D. 0.10 seconds
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   Answer: A. 0.04 seconds

   At 25 mm/sec each 1-mm small box is 0.04 seconds, and each large box (5 small boxes) is 0.20 seconds.

2. At standard paper speed, how much time does one large box (five small boxes) represent?

   1. A. 0.04 seconds
   2. B. 0.20 seconds
   3. C. 0.40 seconds
   4. D. 1.00 second
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   Answer: B. 0.20 seconds

   Five small boxes of 0.04 seconds each equal 0.20 seconds. Five large boxes equal 1 second, and 30 large boxes equal 6 seconds.

3. At standard calibration, each 1-mm small box measured vertically represents:

   1. A. 0.1 mV
   2. B. 1.0 mV
   3. C. 0.5 mV
   4. D. 0.01 mV
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   Answer: A. 0.1 mV

   Standard calibration is 10 mm per 1 mV, so each small box vertically is 0.1 mV. ST elevation is usually described in millimeters at this calibration.

4. What is the normal duration of the PR interval in an adult?

   1. A. 0.12 to 0.20 seconds
   2. B. 0.04 to 0.10 seconds
   3. C. 0.20 to 0.28 seconds
   4. D. 0.08 to 0.12 seconds
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   Answer: A. 0.12 to 0.20 seconds

   A normal PR interval is 0.12 to 0.20 seconds (3 to 5 small boxes). A PR longer than 0.20 seconds defines first-degree AV block.

5. A QRS complex is considered wide when its duration is:

   1. A. 0.12 seconds or longer
   2. B. 0.08 seconds or longer
   3. C. 0.10 seconds or longer
   4. D. 0.20 seconds or longer
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   Answer: A. 0.12 seconds or longer

   A QRS of 0.12 seconds (3 small boxes) or longer is considered wide, suggesting a ventricular origin or aberrant conduction such as a bundle branch block.

6. On a new 12-lead, lead I shows an inverted P wave, QRS and T wave, and aVR is upright. The patient is well and was upright on yesterday's ECG. What is the most likely cause?

   1. A. The patient has had an acute lateral infarction
   2. B. The left and right arm electrodes are swapped
   3. C. The paper speed was set to 50 mm per second
   4. D. The chest leads were placed one space too low
   Show answer

   Answer: B. The left and right arm electrodes are swapped

   Swapping the arm electrodes flips lead I (and switches aVR with aVL), so lead I becomes negative throughout while aVR, normally negative, turns upright. Recheck the electrodes before interpreting. A lateral MI does not invert the P wave, paper speed changes widths rather than polarity, and chest lead position does not affect lead I.

7. The calibration mark at the start of a 12-lead is only 5 mm tall instead of the usual 10 mm. How does this affect what you see?

   1. A. Heart rates appear about twice as fast
   2. B. QRS complexes look twice as wide as they are
   3. C. ST elevation looks about half its true size
   4. D. Voltage measurements are not affected
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   Answer: C. ST elevation looks about half its true size

   A half-standard setting (5 mm per mV) shrinks every voltage by half, so 2 mm of true ST elevation looks like 1 mm and a STEMI can be missed; always check the calibration mark. Calibration does not change the horizontal (time) scale, so heart rates and QRS widths are unaffected, while voltage measurements clearly are affected.

8. Why is atrial repolarization not usually visible on the ECG?

   1. A. It occurs before the P wave begins
   2. B. It produces no electrical current
   3. C. It is hidden within the QRS complex
   4. D. It is shown by the U wave instead
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   Answer: C. It is hidden within the QRS complex

   Atrial repolarization happens at the same time as ventricular depolarization, so the much larger QRS complex obscures it.

9. In lead II, the positive electrode is placed on the:

   1. A. Left arm
   2. B. Right arm
   3. C. Right leg
   4. D. Left leg
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   Answer: D. Left leg

   Lead II runs from the right arm (negative) to the left leg (positive), following the normal direction of depolarization. This produces upright P waves and QRS complexes in a normal heart.

10. Lead I records the electrical difference between which two electrodes?

    1. A. Left arm negative, left leg positive
    2. B. Right arm negative, left leg positive
    3. C. Right arm negative, left arm positive
    4. D. Left leg negative, right arm positive
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    Answer: C. Right arm negative, left arm positive

    Lead I is right arm (negative) to left arm (positive) and views the heart laterally. Lead II is right arm to left leg, and lead III is left arm to left leg.

11. Lead III records the electrical difference between which two electrodes?

    1. A. Right arm negative, left arm positive
    2. B. Right arm negative, left leg positive
    3. C. Left arm negative, left leg positive
    4. D. Left leg negative, left arm positive
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    Answer: C. Left arm negative, left leg positive

    Lead III uses the left arm as negative and the left leg as positive. It is one of the inferior leads along with II and aVF.

12. Which leads are the augmented limb leads?

    1. A. I, II, and III
    2. B. V1, V2, and V3
    3. C. V4R, V5R, and V6R
    4. D. aVR, aVL, and aVF
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    Answer: D. aVR, aVL, and aVF

    The augmented leads aVR, aVL, and aVF are unipolar leads that, together with I, II, and III, make up the six limb leads viewing the frontal plane.

13. Where is lead V1 placed?

    1. A. Fourth intercostal space, left sternal border
    2. B. Fourth intercostal space, right sternal border
    3. C. Fifth intercostal space, right midclavicular line
    4. D. Second intercostal space, right sternal border
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    Answer: B. Fourth intercostal space, right sternal border

    V1 is placed at the fourth intercostal space just right of the sternum, and V2 at the fourth intercostal space just left of the sternum. Placing them too high is a common error.

14. Where is lead V4 placed?

    1. A. Fourth intercostal space, midclavicular line
    2. B. Fifth intercostal space, midclavicular line
    3. C. Fifth intercostal space, anterior axillary line
    4. D. Fifth intercostal space, midaxillary line
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    Answer: B. Fifth intercostal space, midclavicular line

    V4 goes in the fifth intercostal space at the left midclavicular line. V3 is placed midway between V2 and V4.

15. Where is lead V3 placed?

    1. A. Directly over the sternum
    2. B. Midway between V4 and V6
    3. C. Under the left nipple
    4. D. Midway between V2 and V4
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    Answer: D. Midway between V2 and V4

    V3 is placed halfway between V2 and V4, so V4 is usually positioned before V3. Placement by nipple position is unreliable, especially in women.

16. Where is lead V6 placed?

    1. A. Left midaxillary line, level with V4
    2. B. Left anterior axillary line, level with V4
    3. C. Left posterior axillary line, level with V4
    4. D. Left midaxillary line, sixth intercostal space
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    Answer: A. Left midaxillary line, level with V4

    V5 is at the anterior axillary line and V6 at the midaxillary line, both on the same horizontal level as V4. V6 is not simply placed in the sixth intercostal space.

17. To obtain right-sided lead V4R, where is the electrode placed?

    1. A. Left fifth intercostal space, midclavicular line
    2. B. Right fourth intercostal space, sternal border
    3. C. Right fifth intercostal space, midaxillary line
    4. D. Right fifth intercostal space, midclavicular line
    Show answer

    Answer: D. Right fifth intercostal space, midclavicular line

    V4R mirrors V4 on the right side of the chest: fifth intercostal space, right midclavicular line. It is the most useful single lead for detecting right ventricular infarction.

18. When obtaining posterior leads, V8 is placed at the:

    1. A. Right midscapular line, level with V6
    2. B. Left posterior axillary line, level with V4
    3. C. Left paraspinal border, level with V4
    4. D. Left midscapular line, level with V6
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    Answer: D. Left midscapular line, level with V6

    Posterior leads lie on the same horizontal plane as V6: V7 at the left posterior axillary line, V8 at the left midscapular line, and V9 at the left paraspinal border.

19. A regular rhythm has 3 large boxes between consecutive R waves. What is the heart rate?

    1. A. 75 beats/min
    2. B. 100 beats/min
    3. C. 150 beats/min
    4. D. 60 beats/min
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    Answer: B. 100 beats/min

    Using the 300 method, divide 300 by the number of large boxes between R waves: 300 ÷ 3 = 100. The sequence 300, 150, 100, 75, 60, 50 can be memorized.

20. A regular rhythm has 25 small boxes between R waves. Using the 1,500 method, what is the heart rate?

    1. A. 50 beats/min
    2. B. 60 beats/min
    3. C. 75 beats/min
    4. D. 100 beats/min
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    Answer: B. 60 beats/min

    Divide 1,500 by the number of small boxes between R waves: 1,500 ÷ 25 = 60. This is the most precise method for regular rhythms.

21. Which method of calculating heart rate is most appropriate for an irregular rhythm?

    1. A. Divide 300 by large boxes between R waves
    2. B. Divide 1,500 by small boxes between R waves
    3. C. Count QRS complexes in 6 seconds × 10
    4. D. Measure one R-R interval and convert it
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    Answer: C. Count QRS complexes in 6 seconds × 10

    Methods based on a single R-R interval are inaccurate in irregular rhythms. Counting complexes in a 6-second strip and multiplying by 10 gives a reasonable average rate.

22. A patient who is shivering in the cold has a monitor showing an erratic, jittery baseline. The most likely cause is:

    1. A. Somatic muscle tremor artifact
    2. B. Coarse ventricular fibrillation
    3. C. Atrial flutter waves
    4. D. Electrical AC interference
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    Answer: A. Somatic muscle tremor artifact

    Muscle tremors from shivering, seizures, or movement produce irregular baseline artifact. Always check the patient: a talking patient with a pulse does not have VF.

23. The monitor shows a thick baseline of fine, evenly spaced, regular oscillations in all leads near a running medical device. This is most likely:

    1. A. 60-cycle (AC) interference
    2. B. Fine ventricular fibrillation
    3. C. Atrial fibrillation waves
    4. D. Patient muscle tremor
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    Answer: A. 60-cycle (AC) interference

    AC (60-Hz) interference produces regular, uniform fine oscillations from nearby electrical equipment or poor grounding. Muscle artifact is irregular, and fibrillation waves are not perfectly uniform.

24. Why is lead II commonly used for continuous cardiac rhythm monitoring?

    1. A. It is the best lead to detect STEMI
    2. B. It views the posterior wall directly
    3. C. It usually shows clear upright P waves
    4. D. It needs only two electrodes placed
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    Answer: C. It usually shows clear upright P waves

    Lead II parallels the normal flow of depolarization, so P waves and QRS complexes are usually upright and easy to identify. A single lead cannot be used to diagnose STEMI.

25. You are applying a 12-lead to a diaphoretic man with a hairy chest, and the electrodes keep peeling off. What is the best step?

    1. A. Add more gel to the old electrodes
    2. B. Dry the skin and clip hair at the sites
    3. C. Shift the chest leads to drier nearby sites
    4. D. Skip the precordial leads entirely
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    Answer: B. Dry the skin and clip hair at the sites

    Good electrode contact requires dry, hair-free skin; dry the sites and clip hair as needed. Moving precordial leads changes the views and can create false findings. Extra gel does not make old electrodes stick to wet, hairy skin, and skipping the chest leads loses the views needed to find a STEMI.

26. The QT interval represents:

    1. A. Atrial depolarization and AV delay
    2. B. Ventricular repolarization only
    3. C. Ventricular depolarization and repolarization
    4. D. Conduction through the bundle branches
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    Answer: C. Ventricular depolarization and repolarization

    The QT interval runs from the start of the QRS to the end of the T wave, covering ventricular depolarization and repolarization. A prolonged QT increases the risk of torsades de pointes.

27. The J point on the ECG is the:

    1. A. Start of the P wave and the PR interval
    2. B. Peak of the T wave in the precordial leads
    3. C. End of the T wave and start of the U wave
    4. D. End of the QRS and start of the ST segment
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    Answer: D. End of the QRS and start of the ST segment

    The J point is where the QRS complex ends and the ST segment begins. ST elevation and depression are measured at the J point.

28. Which leads view the inferior wall of the left ventricle?

    1. A. I, aVL, and V6
    2. B. V1, V2, and V3
    3. C. II, III, and aVF
    4. D. aVR, V1, and V4R
    Show answer

    Answer: C. II, III, and aVF

    Leads II, III, and aVF look at the inferior surface. I, aVL, V5, and V6 are the lateral leads.

29. Which leads view the lateral wall of the left ventricle?

    1. A. II, III, aVF, and V4R
    2. B. I, aVL, V5, and V6
    3. C. V1, V2, V3, and V4
    4. D. aVR, V1, V7, and V8
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    Answer: B. I, aVL, V5, and V6

    The high lateral leads are I and aVL; the low lateral leads are V5 and V6. Together they view the lateral wall supplied mainly by the circumflex artery.

30. Leads V1 and V2 primarily view which area of the heart?

    1. A. Inferior left ventricle
    2. B. Lateral left ventricle
    3. C. Interventricular septum
    4. D. Posterior left ventricle
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    Answer: C. Interventricular septum

    V1 and V2 are the septal leads, V3 and V4 the anterior leads. Together V1 to V4 view the anteroseptal region supplied by the LAD.

31. The six limb leads view the heart in which plane?

    1. A. Horizontal plane
    2. B. Frontal plane
    3. C. Sagittal plane
    4. D. Oblique axial plane
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    Answer: B. Frontal plane

    Limb leads (I, II, III, aVR, aVL, aVF) view the heart in the frontal (vertical) plane. The precordial leads view it in the horizontal plane.

32. A patient with chest pain shows normal sinus rhythm on 3-lead monitoring. Why is this insufficient to rule out a STEMI?

    1. A. A single lead views only part of the heart
    2. B. The monitor filters out all ST changes
    3. C. Lead II is unable to view the inferior wall
    4. D. ST changes take hours to appear
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    Answer: A. A single lead views only part of the heart

    One monitoring lead sees only a limited area of the heart and may use filtering that distorts ST segments. A diagnostic 12-lead is required to evaluate for STEMI. Lead II does view the inferior wall, but only that area, and ST changes can appear within minutes.