# simple ohms law question

Discussion in 'General Electronics Chat' started by dthx, Jun 14, 2013.

1. ### dthx Thread Starter Member

May 2, 2013
194
14
Would someone please show me...using ohms law...
how voltage drops when resistance increases....
I cant find a thread on it ....
I think Im on the wrong line of thinking....
Coming from a fluid power background from years ago....
I think its confusing me....
If you have a variable oriface in a circuit (variable resistor) and you make it smaller (increase the resistance) .....
given the same flow rate...circuit pressure (voltage) goes UP.....
and yet ...what Im hearing is that elctronically with increased resistance ....voltage goes down.....
Although I hesitate to ask this....I have to .....cause Im stuck...
D.

2. ### blah2222 Well-Known Member

May 3, 2010
554
33
The assumption that you have made is that the flow rate (current in the electrical case) has remained constant.

If there is more resistance in the circuit, this will make it harder for electrons to flow, thus decreasing current. Voltage and current are both related by V = IR.

If the increased resistance decreases the current flow, then there will be less of a voltage drop seen across the resistor.

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3. ### crutschow Expert

Mar 14, 2008
13,003
3,232
Don't confuse voltage with voltage drop. It's the same as absolute pressure versus pressure drop.

Thus if you increase the resistance with the same current, the voltage drop across the resistor goes up. If you increase the resistance with the same voltage drop then the current goes down. This is analogous to fluid flow through an orifice.

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4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Good comment

But it is not the end of the story.

In electrical circuits
(where Ohm's Law rules)

When a load (resistance) is connected across an electrical supply it draws current.

We either choose

A current that is constant regardless of the appled voltage.
( We call this a current cource or cosntant current)

or more commonly
A voltage that is constant regardless of the current drawn.
We call this a voltage source or constant voltage supply.

So if we vary the resistence of the load across a constant voltage supply the current thriought the load will vary.
And if we vary the resistance of a load across a constant current supply the voltage across the laod will vary.

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5. ### dthx Thread Starter Member

May 2, 2013
194
14
Yeah.....youre right....I have made the assumption of constant flow...
Ok....maybe the light is begining to shine, through the haze....
Im gonna take some advice from somewhere....cant remember and go to the board and actually test out a circuit or two....cross checking the values on paper with what my meter says...
Thank you all for a reply....I feel like Im moving in the right direction ...although slowly....
Damnit....this isnt rocket science...what the ^*)*** is so hard about getting my head around the basics...
Wait...its the basics of rocket science....so ...I take that back...

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6. ### #12 Expert

Nov 30, 2010
16,278
6,789
Have you tried that simplistic version I wrote called, "Ohm's Law for Noobies" at the top of the Chat section? It's dead simple, but sometimes that's what a person needs.

After that, do the math for 2 or three fairly simple circuits; a battery and a resistor, a LED, or a light bulb, and that will reveal where you have a false belief. After you strip away the mistaken beliefs, you can progress on a firm, safe, foundation.

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7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Flow rate is analogous to current. Force a constant current through a resistor, and the voltage across the resistor will be proportional to current.
Your confusion may arise from the fact that constant current sources are less common in electronics than constant voltage sources (e.g., batteries, mains voltage, etc.).

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8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
The Ohm's Law triangle is a simple helpmate.

Draw the triangle as shown.

Cover up the wanted or unknown quantity.

Those still visible in the triangle show the appropriate relationship

eg
wanted voltage, cover up V and se that V=IR

wanted current, cover up I and find I=V/R

• ###### ohmslawtriangle.jpg
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