Voltage Drop

[smac999]

Originally Posted by btrayfors

Guys,



Any resistance by definition means voltage drop -- Ohms Law!

V=IR

0 current = 0 voltage drop.

[Wind River]

Simple rules that have served me well:

1. For voltage to drop, current must flow.
2. Voltmeters don't measure voltage, they measure the difference in voltage between the positive and negative leads.
3. Ohmmeters tend to lead you down the wrong path when trouble shooting. They don't give an indication of potential voltage drop. (applies mostly to higher amperage circuits)
4. Voltage drop across the load should equal the source voltage. If there is voltage on the negative side of the load when current is flowing, look for resistance (bad connections) in the ground circuit.

[btrayfors]

smac,

You've just proved my point: whenever R is greater than zero, there's a voltage drop, no matter the I (current).

Moreover, there are voltage drops associated with EVERY SINGLE COMPONENT IN A CIRCUIT, since every component has some resistance! No way around that.

Wind River,

Congratulations. Your "Simple rules that have served me well" are 100% WRONG. All four of them.

That's a new record, I think :-)

Bill

[Wind River]

Originally Posted by btrayfors

smac,


Wind River,

Congratulations. Your "Simple rules that have served me well" are 100% WRONG. All four of them.

That's a new record, I think :-)

Bill

Enlighten me.

[btrayfors]

Originally Posted by Wind River

Enlighten me.

OK, I'll try. One more time.

1. For voltage to drop, current must flow.

Not exactly. Whenever there is a resistance in a DC circuit -- no matter the reason -- then there will be a voltage drop. Most DC circuits are an amalgam of voltage drops occasioned by every single component in that circuit. Moreover, the amount of voltage drop varies according to the amperage flowing in the circuit.


2. Voltmeters don't measure voltage, they measure the difference in voltage between the positive and negative leads.

That's a meaningless distinction. Voltage is the difference in potential between the positive side of any component in a circuit and the negative side of any component in a circuit. BTW, for giggles remember that current generally flows from negative to positive, not the other way :-)


3. Ohmmeters tend to lead you down the wrong path when trouble shooting. They don't give an indication of potential voltage drop. (applies mostly to higher amperage circuits)

Well, yes and no. See #1 above, where voltage drop varies with amperage.


4. Voltage drop across the load should equal the source voltage. If there is voltage on the negative side of the load when current is flowing, look for resistance (bad connections) in the ground circuit.

I think you're referring to Kierkoff's Voltage Law, i.e.,

Kirchoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be equal to zero. This idea by Kirchoff is known as the Conservation of Energy.

Bill

[Wind River]

Originally Posted by Wind River

Simple rules that have served me well:

1. For voltage to drop, current must flow.
2. Voltmeters don't measure voltage, they measure the difference in voltage between the positive and negative leads.
3. Ohmmeters tend to lead you down the wrong path when trouble shooting. They don't give an indication of potential voltage drop. (applies mostly to higher amperage circuits)
4. Voltage drop across the load should equal the source voltage. If there is voltage on the negative side of the load when current is flowing, look for resistance (bad connections) in the ground circuit.

1. Lets take a starting circuit here. To measure the voltage drop on a starter cable from the battery to the starter post. I would put a lead on the battery terminal and one on the starter and crank the starter (current flowing). If there is a voltage drop on that cable it will show up only when the current flows. Also at this point the voltmeter is measuring the difference in voltage between the two points and not the voltage of the circuit (application of rule #2)

3. If this same starter cable is measure for resistance to determine a no start condition and all but a few of the strands of wire in the cable are broken, an ohmmeter may read no resistance but the capacity of the cable to carry the current is compromised, resulting an in large voltage drop. (seen by test #1)

4. Take a simple circuit of a 12 volt source and a light bulb. Measuring from battery positive to negative side of the bulb (while it is lighted) will equal the source voltage. If measured from the negative side of the lit bulb to ground there will be no voltage left unless there is a bad ground and then there will be a voltage differential at that point.

How is this wrong?

[btrayfors]

Originally Posted by Wind River

1. Lets take a starting circuit here. To measure the voltage drop on a starter cable from the battery to the starter post. I would put a lead on the battery terminal and one on the starter and crank the starter (current flowing). If there is a voltage drop on that cable it will show up only when the current flows. Also at this point the voltmeter is measuring the difference in voltage between the two points and not the voltage of the circuit (application of rule #2)

3. If this same starter cable is measure for resistance to determine a no start condition and all but a few of the strands of wire in the cable are broken, an ohmmeter may read no resistance but the capacity of the cable to carry the current is compromised, resulting an in large voltage drop. (seen by test #1)

4. Take a simple circuit of a 12 volt source and a light bulb. Measuring from battery positive to negative side of the bulb (while it is lighted) will equal the source voltage. If measured from the negative side of the lit bulb to ground there will be no voltage left unless there is a bad ground and then there will be a voltage differential at that point.

How is this wrong?

Semantics, only.

I agree with the above scenario. The voltage drop measured between the battery post and the starter (presumably the large post on the solenoid) can be thought of, arguably, as two points on the positive side of the circuit.

The point re: load carrying ability of the starter cable -- possibly compromised -- and resistance measurements is valid, and is exactly what I've repeatedly said.

I think we mostly differ in definitions.

However, I contend that all DC circuits have resistance associated with EVERY single component -- wire, connections and connectors, switches, etc., etc. If you prefer to call these POTENTIAL voltage drop items, OK.

How much the voltage drop will actually be when the circuit is active will depend on the load.

Bill

[Wind River]

Originally Posted by btrayfors

OK, I'll try. One more time.

1. For voltage to drop, current must flow.

Not exactly. Whenever there is a resistance in a DC circuit -- no matter the reason -- then there will be a voltage drop. Most DC circuits are an amalgam of voltage drops occasioned by every single component in that circuit. Moreover, the amount of voltage drop varies according to the amperage flowing in the circuit.

Yes, but in practical trouble shooting at a level most deal with here we are not looking for a .005 voltage drop that these components produce, So when the current is flowing is when most trouble will be found. Many are not aware of the free electron theory of current flow and I think it just confuses the average boater.


2. Voltmeters don't measure voltage, they measure the difference in voltage between the positive and negative leads.

That's a meaningless distinction. Voltage is the difference in potential between the positive side of any component in a circuit and the negative side of any component in a circuit. BTW, for giggles remember that current generally flows from negative to positive, not the other way :-)

I think this has a lot of meaning. It helps visualize what one is looking at when trouble shooting. I once quickly found a bad connection at a fuse on an auto pilot by knowing that if there was a good connection across this fuse there should be no difference in voltage from one side of the fuse to the other, not the 6 volts I had

3. Ohmmeters tend to lead you down the wrong path when trouble shooting. They don't give an indication of potential voltage drop. (applies mostly to higher amperage circuits)

Well, yes and no. See #1 above, where voltage drop varies with amperage.


You made my point here.

4. Voltage drop across the load should equal the source voltage. If there is voltage on the negative side of the load when current is flowing, look for resistance (bad connections) in the ground circuit.

I think you're referring to Kierkoff's Voltage Law, i.e.,

Kirchoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be equal to zero. This idea by Kirchoff is known as the Conservation of Energy.

Bill
The way the law you state reads is my exact point. I'm not sure of your explanation of it though.

Bill
You are obviously miles ahead of me in electrical theory but in practical trouble shooting on a boat think I am far from 100% wrong on all points. If we were discussing electronic component function inside a complicated circuit maybe my assertions will not apply as well.

Thanks for making me think today.

John

[Stu Jackson]

Folks, sounds like we're dancing on the head of a pin again. The OP should measure the voltage at the beginning of the wire and then the end of the wire to figure out the vd. 12.8 V at the beginning, 12.0V at the other end, 0.8V vd. What's so hard?

ITWMB, that's what I'd do. I'd also take a brand new wire, about the same length or maybe even longer to make the temporary replacement new wire simple to temporarily install and do it again. If the one end is 12.8V and the other end is 12.75V, the old wire is toast. Right?

[Wind River]

Originally Posted by Stu Jackson

Folks, sounds like we're dancing on the head of a pin again. The OP should measure the voltage at the beginning of the wire and then the end of the wire to figure out the vd. 12.8 V at the beginning, 12.0V at the other end, 0.8V vd. What's so hard?

ITWMB, that's what I'd do. I'd also take a brand new wire, about the same length or maybe even longer to make the temporary replacement new wire simple to temporarily install and do it again. If the one end is 12.8V and the other end is 12.75V, the old wire is toast. Right?

Sorry, I got called out, and it stung a little.

[model 10]

In addition to all the other junk we carry on board, we should carry a meg ohm meter, then that pesky battery cable would show it's true nature.

[Rusty123]

Originally Posted by smac999

V=IR

0 current = 0 voltage drop.

Smac, you are correct in theory. Problem is that there is no way to measure voltage without introducing current (from the meter), which makes the scenario deviate from theory.

[Rusty123]

Originally Posted by btrayfors

Semantics, only...If you prefer to call these POTENTIAL voltage drop items, OK.

Bill, I think this resolves much of the disagreement in the thread. That is, all components have resistance, and therefore will potentially cause voltage drop when current flows. But until current flows, the potential throughout the connected component will be equal, in theory.

Of course, the problem is that there is no way to measure that potential without connecting a meter, which introduces current, which blows the theoretical predicted values out of the water.

[fryewe]

Even the small voltmeter current through a high resistance will cause voltage drop and such measurements are one effective way to check for high resistance... a milliamp flowing through a kilohm drops the voltage by a volt...as does a microamp flowing through a megohm.

Went through a similar troubleshooting on my tractor just yesterday.

Wouldn't start. Battery voltage was 12.6 measured on the posts. Was 12.0 measured on the cable end terminals. Cleaned the positive end terminal and post. Measured again and terminal voltage was 12.6.

Tractor still wouldn't start. Measured the voltage at the starter interlock relay and found it to be 8.5 volts. Removed the relay and fuse panel (which was dusty and dirty...recurring issue on a tractor) and blew it out with dry compressed air. Measured voltage went up to 9.5 volts.

Tractor started.

The OP likely has a high resistance connection on the wire in question. Stu's (and others) recommendation to run a parallel wire and check the V drop on that wire will likely confirm it (27 feet!).

If there are multiple loads with such long runs of wiring to that location perhaps it would be better to run a ten gauge wire to the area and terminate it at a terminal strip to provide power to those loads through shorter small gauge wiring.

[Terra Nova]

Originally Posted by fryewe

...If there are multiple loads with such long runs of wiring to that location perhaps it would be better to run a ten gauge wire to the area and terminate it at a terminal strip to provide power to those loads through shorter small gauge wiring.

One 10-gauge (duplex) wire won't carry much over 27 feet--not for supplying a sub-panel. Better to run everything from the main panel.