What exactly Is voltage drop and how Is It detected?

[Reprovo]

Very clear explanation.Thanks!

 

[Arouse1973]

Yes.

Say we have a 9V battery (let's assume it's a perfect voltage source) with a 100k (kilohm) resistor connected across it. The current that will flow through the resistor can be calculated using Ohm's Law: I = V / R where V is 9V and R is 100,000 ohms. 9 / 100,000 is 0.00009 amps, or 90 µA (microamps).

If you connect a voltmeter across the resistor, you now have a 100k resistor and the voltmeter, in parallel, connected across the battery. The voltage across each component is 9V. The current through the resistor doesn't change. Current will flow through the voltmeter. If it's a digital multimeter with a 10 megohm input resistance, this current will be 0.9 µA (by Ohm's Law). So the total current coming out of the battery will be 90.9 µA.

If the voltmeter is an analogue one (I'm mentioning this because you mentioned "deflection"), and has an input impedance of 20 kilohms per volt, and is set to the 10V range, it will have an input impedance of 200 kilohms. With 9V across it, 45 µA will flow through it, and the total current drawn from the battery will be 135 µA.

Now, if we break the circuit and connect a voltmeter in SERIES - in other words, battery positive, through resistor, through voltmeter, back to battery negative, the 9V from the battery is applied across the series combination of the resistor and the voltmeter, and will be divided according to the relative resistances of the resistor and the voltmeter.

If it's a digital multimeter, the total resistance (resistors in series add together) will be 10.1 megohms. Therefore the current will be 9 / 10,100,000 which is 0.89109 µA. This current flows though both resistances, and the voltage across the resistances can be calculated from V = I R. For the 100k resistor, V will be 0.089109 volts, and for the multimeter, V will be 8.9109 volts. So the mutimeter will read 8.9109 volts.

You can say that there is 9V coming from the battery, and a voltage drop of 0.089109 volts across the resistor, so the remaining voltage, 8.9109 volts, appears across the multimeter.

If it's the analogue voltmeter, the total resistance will be 300 kilohms. The current will be 30 µA and the voltages across the resistor and the voltmeter will be 3V and 6V respectively. These voltages are in proportion to their resistances. So there will be a voltage drop of 3V across the resistor, and the analogue voltmeter will read 6V.

Kris
What a very detailed explanation as always. But I thought you said you should use the term in series with the battery not across the battery.
Thanks
Adam

 

[KrisBlueNZ]

What I said was that a current source can be made with a resistor in series with a battery. This circuit has two output terminals and if you connect a load across them, current will flow through it.

A circuit consisting of a resistor connected across a battery also has two connection points, and if you connect a (another) load across them, it will see a voltage, so it's not appropriate to describe it as a current source.

If you want to describe a resistor connected across a battery as a current source, you have to break the connection somewhere to insert the load. The current source part of the circuit then becomes a resistor in series with the battery.