# Basic Electronics

Discussion in 'General Electronics Discussion' started by Akshatha Venkatesh, Mar 17, 2017.

1. ### Akshatha Venkatesh

145
0
Jan 14, 2017
hi, Can anyone help me with the circuit below.I just trying to understand the concept of voltage drop across a resistor. what will be the voltage at point A and B in the figure I've attached below.
Thank you.

2. ### Externet

785
169
Aug 24, 2009
If you apply voltmeter probes to points A and B, the voltage read across that resistor will be 10 Ω times the current in Amperes flowing trough the resistor.
If you do not apply the probes to the circuit but just want to calculate the result, the answer is the same, resistance times current.

3. ### Akshatha Venkatesh

145
0
Jan 14, 2017
I want the volatge at point A , not across the second resistor. Will the voltsge at point A be (6V-drop across R562)?

4. ### Externet

785
169
Aug 24, 2009
You said first "the voltage drop across a resistor"

Now, the voltage at point A in reference to what ?
If the positive probe of your voltmeter is sensing A; where in your circuit is the negative probe of your volt meter ?

5. ### Akshatha Venkatesh

145
0
Jan 14, 2017
With reference to ground

145
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Jan 14, 2017

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7. ### BobK

7,682
1,688
Jan 5, 2010
6V. The ADC inputs of the pull so little current than even with 10K impedance the voltage drop will be less then the resolution of the ADC. So with only 572Ω, the drop will not be measurable on an ordinary multimeter.

Bob

8. ### AnalogKid

2,544
725
Jun 10, 2015
Your circuit actually has three resistors in series. The first two are the two you show. The third is the effective input impedance of the A/D converter. This impedance is to GND, and completes the input circuit. Hopefully it is very high, because if the input draws current through the external resistors, there will be voltage drops across those resistors and the A/D will not see the full 6 V (or whatever the input voltage is).

There are many different kinds of A/D methods, and some of them have an input impedance that is low enough to cause errors. Others have an input impedance that changes depending on the frequency of the signal.

ak

9. ### Externet

785
169
Aug 24, 2009
There is no ground on your first schematic.

Now, there is no equivalent load or resistance to ground at the ADC pin. And you do not know it.

If the equivalent internal resistance at such MCU pin was 1 MΩ to ground, the current flowing through the resistors is
6V divided by 1000572Ω = current

That current multiplied by 562Ω gives the voltage across the 562Ω resistor.

Substracting that voltage across the 562Ω resistor from 6 V gives you the voltage at A in reference to ground.

10. ### BobK

7,682
1,688
Jan 5, 2010
Which, in the example of 1MΩ, would be make the voltage 5.996V, or 6V to 3 digit accuracy. And the input current ot the ADC is probably even less than that in reality.

Bob

11. ### hevans1944Hop - AC8NS

4,631
2,159
Jun 21, 2012
Voltage is never measured AT a point. It is always measured BETWEEN two points. The voltage drop across a resistor is given by Ohm's Law to be the product of the current through the resistor and the resistance value:

V = (I)(R), where V is measured in volts, I is measured in amperes, and R is measured in ohms.

You have labeled "Votage at this point = 6V" but, as others have stated, this is meaningless. Then you later said this is measured with respect to "ground" but do not show on the schematic where the "ground" is. As it stands, without knowing what is on the remainder of the schematic, where you have written "This point goes to the ADC pin of a mcu" there is no way to determine what current (if any) is present in the resistors. If there is a current present, it will be the same current in both resistors because they are connected in series. This is a consequence of Kirchoff's Current Law: Whatever currents enter a node must also leave that node, paying close attention to algebraic signs. Or stated another way, the algebraic sum of all the currents entering or leaving a node must be zero. A node is any point on a schematic diagram to which two or more wires are attached. The red points on your diagram, two of which are labeled "A" and "B" are nodes.

With respect to "ground" the ADC input represents an additional resistance in series with the two you have drawn and that resistance, along with the "6V" input (with respect to "ground") will determine the current through both resistors and the ADC input. That current will be 6V divided by the sum of all three resistances, the two you have shown plus the ADC input resistance. Some ADCs on MCUs have relatively low input resistances, on the order of a few thousand ohms, which can affect the accuracy with which you can measure the "6V" input because the current drawn by these low input-resistance ADCs causes a voltage drop across the other resistances in series with the ADC input. Read the datasheet for the MCU to find out what the ADC input resistance is and then calculate what effect that has on your input voltage measurement.

12. ### Externet

785
169
Aug 24, 2009
Perhaps this will help too :

(+)10V----------50Ω--------A-------30Ω------B------20Ω-------C-----------(-)

Must know the current : 10V divided by 100Ω = Current is 0.1A
Voltage dropped by 50Ω = 50 x 0.1 = 5V = between (+) and A
Voltage dropped by 30Ω = 30 x 0.1 = 3V = between A and B
Voltage dropped by 20Ω = 20 x 0.1 = 2V = between B and C
Voltage at A respect to (-) = 5V
Voltage at B respect to (-) = 2V
Voltage at C respect to (-) = 0V

So, if you have 10V and need 2V; connect to B and (-)

13. ### duke37

5,364
772
Jan 9, 2011
Ohms law V = I * R

If you were to connect B to ground, then the current would be 6/572 A
The voltage at point A relative to ground would therefore be 60/572 V
In other words a simple voltage divider.