Hi all I'm new here and very new at electronics. For years I wanted to learn some basic electronic circuit but never got the chance to get around it.

This is a very basic question but this is my first try at electronics!!!

My voltage source is 9VDC

If I put a 1.8Mohm across I get 8.5V
If I put a 150ohm across I get 7.6V

I thought higher the resistance lower the voltage!!!

 

Use the correct formula's to measure the voltage drop.

I found this information and thought it would be helpful

How to calculate voltage drop across series resistors.

Step 1: Determine Current of total resistors across the series.
(I = Current/V=Voltage/R(t)=Resistance Total)
I = V / R(t)
I = 9 / (2k + 5k + 10K)
I = .53 mA

Step 2: Now that we have the Current across the resistors we can calculate the voltage across EACH resistor.
I = Current
Vx=Voltage (x=Current Value across X Resistor)
Rx=Resistor value(x=resistor#)

V = I x Rx

V1 = .53 mA x 2k
V1 = 1.06V

V2 = .53 mA x 5k
V2 = 2.65

V3 = .53 mA x 10k
V3 = 5.3V

V1 + V2 + V3 = 9.01 (voltage is 9V only 9.01 because i rounded up from .529 to .53) At .529 you can see barely a difference:

V1=1.058V
V2=2.645V
V3=5.290V
V Total= 8.993V

Also If the resistors are all the same value its easier.

So just add them up:

I = V / R(t)
I = 9 / (5k + 5k + 5K)
I = .6 mA

THEN:


V = I x Rx

V1 = .6 mA x 5k
V1 = 3V
so since all Resistors are same value:
V2 = 3V
V3 = 3V

Also To get different values for like switches just use gnd as a start point and the end of each resistor and a end point. This way you get 3 switches from this each having different voltages

SW1 = 3V
SW2 = 6V
SW3 = 9V

Just remember that with your resistors the voltage drop will be minimal. Since you only have one resistor in line you will most likely see a higher increase in current with the larger value resistor and a lower drop in voltage.

So I hope someone can learn something from this.

Good luck

 

Hi Electronewb! Welcome to AAC!

Resistance does not reduce voltage. It reduces current. Voltage, current, and resistance is related as shown as Ohm's Law: V=I*R, where V is voltage in volts, I is current in amps, and R is resistance in ohms. When you put a 150 ohm resistor across a 9 volt battery, the current through it will be 9=I*150 = 0.06 amps (60 milliamps). When you put a 1.8 megohm resistor across 9 volts, the current will be 9=I*1,800,000 = 0.000005 amps (5 microamps). The higher the resistance, the lower the current. The voltage drop across a single resistor connected to a battery will always be the full voltage of the battery.

 

The result also indicates the battery has some internal resistance. Can you work out what the value of the internal resistance is?

 

Ah OK thanks guys I think I get it. I will have to wrap my head around Ohm's law!!! I guess what I had in mind was a voltage regulator!!! I really thought that a resistor could decrease 10V to let's say 2V. Might be a dumb question but why the use of a resistor? I know it creates a path for electrons to flow but why always reducing the current?

 

Ohm's law and the specific questions you're asking are well explained in Chapter I of the ebook linked at the top of this page and by the experiments linked below.

http://www.allaboutcircuits.com/vol_6/chpt_2/5.html