Im a very beginner trying to learn some electronics basics. I am trying with resistors. I connected a resistor to a 12v power supply in series and tested the voltage using multimeter. But it keeps the same as the actual volage. No voltage drop shows with or without resistor. What could be the reason?

  • \$\begingroup\$ I think you mean the resistor was in parallel with the supply. \$\endgroup\$ – Martin Jul 12 '17 at 11:06
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    \$\begingroup\$ No.. I connected the resistor to the negative of the power suply and tested the voltage by one lead of the multimeter to resistor and the other to the positive of the supply \$\endgroup\$ – Sandeep Thomas Jul 12 '17 at 11:09
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    \$\begingroup\$ @Martin: With just a power supply and a resistor, the resistor is both in series with and parallel to the power supply. With only two two-leaded parts, there is no parallel/series distinction. \$\endgroup\$ – Olin Lathrop Jul 12 '17 at 11:23
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    \$\begingroup\$ Please clarify how you have connected them. As of now there are two equally valid answers matching your description. \$\endgroup\$ – pipe Jul 12 '17 at 11:43
  • \$\begingroup\$ @pipe Yea.. I connected it in series.. \$\endgroup\$ – Sandeep Thomas Jul 12 '17 at 12:23

This is the circuit you described in the question (this site has a built in schematic editor, it's a lot better than trying to describe a circuit using words):


simulate this circuit – Schematic created using CircuitLab

A resistor in the power supply negative terminal and then a meter from the resistor to the power supply positive. So the voltmeter and the resistor are in series. This isn't a normal situation, normally a voltmeter is placed in parallel with the voltage to be measured.

As a first approximation the voltmeter has an infinite resistance. So your total series resistance is therefor infinity+R1 = infinity.

Ohms law states that V = I * R. For the circuit as a whole: V=12, R = infinity so I = 12 / infinity = 0

For the resistor V = I * R, I = 0 (current must be the same at all points in a series circuit) which means that V = 0.

So the voltage drop across the resistor is 0.

If we have 12 V total with 0 V across the resistor the voltage across the meter (the number it will display) will be 12 - 0 = 12 V

In reality the meter will have a finite resistance but it will be in the M Ohm range however for most values of R1 this will be close enough to be considered infinite. If you changed R1 to be close to the value of the meter, something in the 5-10 MOhm range, then the voltage your meter is measuring will start to drop.

  • \$\begingroup\$ Thanks a lot for the detail. Btw Id like to clarify one more thing.. When we connect resistor in series to an LED it should connect to the negative terminal of the LED or positive??? \$\endgroup\$ – Sandeep Thomas Jul 12 '17 at 12:22
  • \$\begingroup\$ It doesn't normally matter. All you are doing is limiting the current that can flow, since current in a series circuit is the same anywhere in the circuit and you don't have any other connections it doesn't matter which order things are in. That said you may find it easier to follow how things work if you put them in a specific order. \$\endgroup\$ – Andrew Jul 12 '17 at 13:55

It seems you have this circuit:

If you measure the voltage on V+ with respect to V-, you should get 12 V. That's basically by definition of what the voltage source does. The voltage source always puts out 12 V.

The resistor causes current to flow. You can think of the resistance deciding how hard the voltage source has to work to maintain the 12 V, but a ideal voltage source will always do that regardless of what resistance you connect to it.

You can calculate the current the voltage source must produce to maintain the 12 V by applying Ohm's law. That says that the current thru a resistor is the voltage applied across it divided by the resistance. In common units:

    A = V / Ω

where A is the current in Amperes, V the EMF in Volts, and Ω the resistance in Ohms.

For example, if R1 is 100 Ω, then the current will be (12 V)/(100 Ω) = 120 mA.

Note that it takes power to maintain a voltage across a resistance. Power is voltage times current. In the example above, the resistor will dissipate (12 V)(120 mA) = 1.44 W. That would blow up typical ¼ and ½ Watt resistors. A 2 W resistor would be fine, but would probably get too hot to touch for long.

Of course real power supplies have real limitations, unlike the theoretical ideal voltage source we were assuming above. Your power supply can only put out so much power, which at a fixed voltage is the same as saying it can only supply so much current.

Let's say your power supply is rated for 1 A. That means it will only maintain its output at 12 V when you try to draw 1 A or less from it. In terms of putting resistors across it, that means it will work with 12 Ω or more. If you put less than 12 Ω across it, meaning you would draw more than 1 A at 12 V, the power supply doesn't guarantee its output will be 12 V anymore. What happens next depends on the power supply. Most will drop their output voltage. Some might shut down for a few seconds, then try again. Others might blow a fuse.

  • \$\begingroup\$ If so how the led on the same circuit gets the voltage drop to 2.2 volt when we connect it along with 1k resistor? \$\endgroup\$ – Sandeep Thomas Jul 12 '17 at 12:32

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