# resistor

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#### andrew8485

##### New Member
As some of you can see from my recent posts I am new to electrontics and I think I learned all about transistors,relays, caps,dilodes , but I am stuck once again. I know the "ohms law" fairly good, I understand a resistor is measured in ohms. Here is the question how do I figure out the volts, and amps that are coming out of the resistor. I think I am mis understanding some thing that I read in a tutorial. thanks once again.

Andrew, there is nothing 'coming out' of a resistor if it sits there all by itself. Its a purely passive element, not at all like a battery.
When you pass a certain current through a resistor of a certain Ohm value you have a certain voltage drop across the resistor. The current passing through the resistor encounters a "resistance", causing it to heat up to some extend. This requires energy, hence there is a voltage drop across the resistor and the resistor dissipates power (Watts).
You calculate the appropriate figures by applying Ohms law which, as you mention, you are familiar with :wink:
I'll all make sense soon enough Yes i understand that you need a power supply, but what I was trying to ask is ,What would the output be in terms of, volts and amps compared to the input Ex. input = XX volts And XX amps With a restistor of XXX ohms will have a output of XX volts And XX amps. I know a resistor reduces the amount of ohms that can pass by, but how do I caculate the amount of volts and amps it lets buy? becuse what I have learned so far is that you need two types of electrical measurements (amps,volts,resistance,watts) to figure the other two measurements out, and I only know the amout of ohms exiting the resistor. thanks

I will use a water analogy to help you understand the relation between volts, ohms and amps. Volts is like the water pressure that forces water thru a hose. Amps is like the flow of water thru the hose. The water meter on your house measures the flow of water, it is like an ammeter. Resistance (ohms) is like the resistance to flow that you would have if the hose was rough inside, or the hose was too small. I hope this helps.

"ohms" cannot pass by. They are the unit of measurement for the resistance that the resistor creates. However, you are correct in that you can use ohms law to calculate the ratio of voltage and current in and out of a resistor.

I didn't understand what you wanted to know about resistance.... I guess Klaus's post is 'wow' in terms of explanation, but if still you need more, or other stuff, please explain a little bit more...

Klaus you got the closest to answering my question by saying "This requires energy, hence there is a voltage drop across the resistor and the resistor dissipates power (Watts).
You calculate the appropriate figures by applying Ohms law which, as you mention, you are familiar with "
but how do you do this can you give me a example pleases thank you

Ohm's law states that the voltage (dropped) across a given load (resistor) is equal to the current flowing in the circuit multiplied by the resistance through the load in question -- E = I x R ...

Assume a 12 volt source and a total series circuit resistance of 100 ohms made up of (1) 47 ohm resistor in series with (1) 53 ohm resistor. The total series circuit current flow is 12 volts divided by 100 ohms, or 0.120 amps. The current flow is constant throughout the series circuit, with the individual drops across each resistor being dependent upon the resistance of each one:

R1 -- 47 ohms -- E = .12 x 47 = 5.64 volts

R2 -- 53 ohms -- E = .12 x 53 = 6.36 volts

Note that the sum of the two individual drops is equal to the source voltage (assuming no wire losses...). The "dropped" voltages can thus be calcualted, and can also be measured by placing a voltmeter across each resistor.

so is what i am about to say right. After the 2 resistors in series (which will equal 100 ohm) the voltage will be 0.00000 I don't think I got what you were trying to say. Sorry for being a pain

andrew8485 said:
so is what i am about to say right. After the 2 resistors in series (which will equal 100 ohm) the voltage will be 0.00000 I don't think I got what you were trying to say. Sorry for being a pain

OK -- here's the circuit that I described:

When measuring voltages at any point in a circuit, the reference must be clearly understood if you wish to obtain meaningful readings. Considering the circuit above, the measurements that I mentioned in my previous reply would be those taken across each resistor, i.e. with the VM probes at point D and E for R1, and at points F and G for R2. If, on the other hand, you are measuring with respect to chassis (ground), the expected voltages would be different -- they would in fact be the source voltage less the accumulated circuit drop to the specific point at which the measurenment is taken.

Watch this -- placing the VM - probe at point A or H and the VM + probe at point C would give an anticipated reading of 12V -- you are in effect measuring the drop across both R1 and R2. Move the + probe to point F, and the anticipated reading is now 6.36V, as you are measuring the drop across R2 only. Move the + probe to point D and the - probe to point E, and the anticipated reading is now 5.64V -- the drop across R1. With the - probe at point H and the + probe at point G, the anticipated reading would be 0V, with any reading other than that indicating an abnormal drop between those points. Similarly, if you connect the VM between points A and H, you should expect a 0V reading, with any other result showing a problem with the circuit.

This measuring technique is useful for locating opens and shorts in a circuit. Suppose the wire were broken between points C and D. What should the voltage read, with respect to ground, at point C? At point D? If the wire were intact, the voltage would be source voltage (12V) at both points. If the wire were broken, Point C would show the source voltage, while point D would show zero volts. In essence, an open is an infinite resistance, with full source voltage being dropped across it. As to locating shorts, consider the drop across R2 if the resistor were somehow shorted. With a direct short, there would be no drop across the ressitor, meaning that (in the sample circuit given) the circuit current would now increase to .255A (12/47), as the entire source voltage would be dropped by R1. The two hints here at a circuit problem are the higher-than-expected current (0.255A vs. 0.120A) and the lack of any voltage drop at R2. If the value of either resistor were to change, its effect on circuit current, and thus voltage dropped across the resistor, will also change.

This technique can thus be used to locate abnormal circuit resistances due to poor soldering, loose connections, undersized conductors, and so on. Each instance of sub-par circuit condition will likely cause a voltage drop at the location of the problem. For example, if you were to measure (with respect to ground) a different voltage at point E than at point F, you would have an indication of a voltage drop along the wire or PCB trace between those two points. While wire losses are always present to a certain extent, a well designed circuit will have minimal (extremely low) wire losses. 