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Attached below is a question from my textbook "Guyton and Hall Textbook of Medical Physiology Review, 4e" and I am confused because the solutions manual has the answer as B (0.5mV), implying that it only took the R wave into consideration for determining the QRS voltage. I remain convinced that the answer should be A (0.05mV) as the Q and S must also be accounted in calculations.

My logic:

R:0.9, Q:-0.15, S:-0.3 = net: 0.45

I+III=II

0.4+I=0.45

What do you think the right answer is? can you please correct me and help me understand the right way of approaching such a question?

Many thanks in advance!

ECG question

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Einthoven’s Law

From ECG and Echo Learning:

This law implies that the sum of the potentials in lead I and lead Ill equals the potentials in lead II. In clinical electrocardiography, this means that the amplitude of, for example, the R-wave in lead Il is equal to the sum of the R-wave amplitudes in lead I and III.

So the law uses the R wave, which is the upward deflection above the isoelectric line.

ECG tracing with coloured waves

Image from Geeky Medics.

Einthoven’s law can be summarised as I + III = II.

Solution

In the ECG you provide, the lead II R-wave has a voltage of 0.9mV. The given value for III is 0.4.

Thus:

I + 0.4 = 0.9

I = 0.9 - 0.4

I = 0.5

So the R-wave voltage in lead I is 0.5mV.


Discussion

Your method of finding the net value of the positive and negative deflections will yield the net deflection, but it is not the net deflection that follows Einthoven’s law.

I think of this as the R wave is the only part of the waveform where the potential difference is positive in the direction of that lead.


Other reference

ECG and Echo Learning

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    Oh I get it now! Thank you so much!! So basically: Mean Electrical Axis: you use net QRS voltage Einthoven’s Law: you use R wave amplitude Nov 25, 2023 at 3:25
  • Yes that seems to be it - It has been years since I studied this but looking at it again that seems to be the way to do it.
    – Chris
    Nov 25, 2023 at 7:31
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    Beautiful. Thank you so much! Nov 25, 2023 at 11:59

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