Electrical Engineering: Basic Laws (25 of 31) The DC Voltmeter

In this video I will explain what is a DC voltmeter.

Next video in this series can be seen at:

[NisseOhlsen]

BOOM! Another ace lecture.

Not only content is good, also for under graduates, but the way you break up your lectures makes them perfect for fast learning.

My complements.

[kallewulf]

Thank you a lot for these amazing lectures!

One point that i was missing in the video was how we would get from the current (measured by the Galvanometer) to the voltage across the branch that we actually want to measure with the DC Voltmeter. For that, simply Ohm's law can be used:

V = I_M * R_G

Where V is the Voltage we want to measure, "I_M" is the current measured be the Galvanometer and "R_G" is the resistance of the resistor connected in series to the Galvanometer. I am assuming the resistance of the Galvanometer itself is so small that it can be neglected.

[choutapelly]

Your a genius!! This thing bugged me for nearly 6 months!! Thank you

[suhailasaba]

2:21
Question: At this moment one might say that R sub G is not in parallel with the Load Resistance as they share only one node.
In that case we can draw the part that contains the galvanometer as R sub G with the very small resistance of the galvanometer in series to show that their R equivalent which is practically almost the same as R sub G is actually in parallel with the Load Resistance. Isn't that right ?

Great thanks for great instructor.

[yellowcyborg]

needed to know how a voltmeter works, and this explained it perfectly! Thank you!

[suhailasaba]

2:10
Question: As it is called R sub G, Does the " sub G " stand for something ; In other words does the letter G stand for something ?

[kangwamulenga]

Y did the current become 2 since the current that was flowing in load resister was 0.02A And RG resister has the same resister of 1000 ohms it also has a current of 0.02A if we sum up the current we are getting 0.04 as our tota or (equivalent) current

[Tahycoon]

if we make the resistance of the voltmeter too high, wouldn't no current (or very little) be crossing the voltmeter and therefore the Galvanometer, which means we will see the needle at or close to 0?