Question

At time t, the cardiac dipole is oriented at - 45° (minus forty five degrees) to the horizontal axis. The magnitude of the dipole is 3 mV. Assuming Einthoven frontal plane configuration, what is the magnitude (in mV) of the electrical signal in lead II? (Round off the answer to two decimal places.)

Hint:

Always remember the standard angles for Einthoven's leads (I: 0°, II: 60°, III: 120°). The measured voltage is always the dipole magnitude multiplied by the cosine of the angle *between* the dipole vector and the lead axis.

Answer 1

Step 1: Understanding the Question:
We are given the magnitude and direction (angle) of the heart's electrical vector (cardiac dipole). We need to find the magnitude of the voltage that would be measured by Lead II in the Einthoven's triangle configuration.
Step 2: Key Formula or Approach:
The voltage measured by any lead in the Einthoven configuration is the projection of the cardiac vector onto the axis of that lead. The formula is: \[ V_{\text{lead}} = V_{\text{dipole}} \times \cos(\theta) \] where \(\theta\) is the angle between the cardiac dipole vector and the lead axis.
Step 3: Detailed Explanation:
The standard angles for the Einthoven leads are:

• Lead I axis: 0°
• Lead II axis: +60°
• Lead III axis: +120°

We are given:

• Magnitude of the cardiac dipole, \(V_{\text{dipole}} = 3 \text{ mV}\)
• Angle of the cardiac dipole, \(\theta_{\text{dipole}} = -45^\circ\)

The lead we are interested in is Lead II, which has an angle \(\theta_{\text{lead II}} = +60^\circ\).
The angle \(\theta\) between the dipole vector and the Lead II axis is the difference between their angles: \[ \theta = \theta_{\text{lead II}} - \theta_{\text{dipole}} = 60^\circ - (-45^\circ) = 105^\circ \] Now, we can calculate the voltage in Lead II: \[ V_{\text{lead II}} = 3 \times \cos(105^\circ) \] We know that \(\cos(105^\circ) = \cos(60^\circ + 45^\circ) = \cos(60^\circ)\cos(45^\circ) - \sin(60^\circ)\sin(45^\circ)\).
Using standard trigonometric values: \[ \cos(105^\circ) = \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2} - \sqrt{6}}{4} \approx -0.2588 \] So, the voltage is: \[ V_{\text{lead II}} = 3 \times (-0.2588) \approx -0.7764 \text{ mV} \] The question asks for the magnitude of the electrical signal. \[ |V_{\text{lead II}}| = |-0.7764| \approx 0.7764 \text{ mV} \] Rounding off to two decimal places, we get 0.78 mV.
Step 4: Final Answer:
The magnitude of the electrical signal in lead II is 0.78 mV.