Copy of 5CL Pre-Lab 7 Submission Template - S23v2

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University of California, Los Angeles *

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5C

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Electrical Engineering

Date

Feb 20, 2024

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pdf

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3

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Q1: The figure shows equipotential lines for a dipole formed by charges whose potentials are +15V and -15V. Suppose you want to measure the potential difference between the points A and B with a voltmeter. Where would you place the positive and negative voltmeter leads? What value of the potential difference will the voltmeter read (both sign and magnitude)? 2 I would place the positive voltmeter lead on +q and the negative voltmeter lead on the -q. The potential difference will read +30V.
3 Q 2: To measure the EMG signal of muscle activation, you can think of the depolarizing cell action potential direction as the direction down your forearm toward your finger. State where you think is a good place to put the sensor stickers such that you can measure a potential difference when the action potential is moving down your forearm (include a sketch). Explain why you choose to arrange the sensors like that. A good place to put the sensor stickers would be on the wrist, with the positive pole on the upper forearm and negative pole on the lower forearm because the wrist is the most dense with nerve fibers and the positive pole should be placed closer to the heart. ~
4 Q 3 : The left figure shows two possible lead placements for measuring EKG signals. The middle figure depicts the heart’s polarization during the R wave, and the arrow indicates the magnitude and direction of the net polarization dipole produced. Which lead placement (Lead I or Lead II) would measure a larger voltage signal during the R wave? Why? What about for the Q wave, as depicted in the figure on the right? Lead 2 will measure a larger voltage signal during the R wave. The positioning of th leads affects the angle at which the heart is viewed. With the second positioning, the it more closely matched the angle of the arrow so the voltage signal will be greater. Lead 1 will measure a larger voltage signal during the Q wave. Similar to the reasoning before, the arrow angle more closely matches the angle that the first lead positioning makes, showing a larger voltage signal.
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