- Location
- Massachusetts
I agree, I would want to remove any doubt in the meter readings.
help with dryer circuit please.
- Thread starter Charlie Bob
- Start date
- Status
I agree, I would want to remove any doubt in the meter readings.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100207-0939 EST
Charlie Bob:
My following comments may seem harsh, but treat your problem as a learning experience.
You have been instructed correctly above to use a large load to test a circuit for problems with poor connections.
Your customer was perfectly correct in telling you to solve the problem at your expense. Your customer went to the extra expense to buy a new dryer because you probably incorrectly diagnosed the problem.
A question without an answer yet --- did the new dryer come with a new cord and plug? This might be critical as to whether the misdiagnose was all your problem. Note: you could have done a resistance check on the neutral conductor in the cord.
Your following statement from the first post is not at all clear:
Note: 110 +117 = 227 is not 245. This should be a red flag.
I have no idea what you meant by " when i check at the dryer is showing open neutral". Checked what, and open neutral where?
Your problem is a good illustration of why an understanding of electrical theory, and how various instruments work is important.
I favor using a high impedance voltmeter and when needed adding a shunt load like a 1500 W heater to check for a high impedance source. But I am not an electrician and a Wiggy as an only tool would be more convenient.
You could have a poor connection, assume 5 ohms, in the neutral. I do not know the impedance of a Wiggy, but I will guess at 1000 ohms. A change of 5 ohms in 1000 ohms at 120 V only produces a voltage change of 120*5/1000 = 0.6 V. By comparison a 1500 W heater when hot is about 10 ohms. A 120 V source with a 5 ohm and 10 resistor in series as the load will produce about 120*10/15 = 80 V across the 10 ohm (heater) load.
Using a modern high end high impedance digital voltmeter should give you 0.1 V resolution on a range reading 120 V. One can not realistically put a continuous 1500 W resistor for 120 V in this meter. You can do a lot of useful troubleshooting or circuit tracing with a meter with 0.1 V resolution, and an external 1500 W load.
.
Charlie Bob:
My following comments may seem harsh, but treat your problem as a learning experience.
You have been instructed correctly above to use a large load to test a circuit for problems with poor connections.
Your customer was perfectly correct in telling you to solve the problem at your expense. Your customer went to the extra expense to buy a new dryer because you probably incorrectly diagnosed the problem.
A question without an answer yet --- did the new dryer come with a new cord and plug? This might be critical as to whether the misdiagnose was all your problem. Note: you could have done a resistance check on the neutral conductor in the cord.
Your following statement from the first post is not at all clear:
I think you were saying --- I disconnected the neutral wire of the cord from the neutral terminal in the dryer. Then I read 245 line to line, 110 from one line to the cord neutral, and 117 from the other line to the cord neutral.
Note: 110 +117 = 227 is not 245. This should be a red flag.
I have no idea what you meant by " when i check at the dryer is showing open neutral". Checked what, and open neutral where?
Your problem is a good illustration of why an understanding of electrical theory, and how various instruments work is important.
I favor using a high impedance voltmeter and when needed adding a shunt load like a 1500 W heater to check for a high impedance source. But I am not an electrician and a Wiggy as an only tool would be more convenient.
You could have a poor connection, assume 5 ohms, in the neutral. I do not know the impedance of a Wiggy, but I will guess at 1000 ohms. A change of 5 ohms in 1000 ohms at 120 V only produces a voltage change of 120*5/1000 = 0.6 V. By comparison a 1500 W heater when hot is about 10 ohms. A 120 V source with a 5 ohm and 10 resistor in series as the load will produce about 120*10/15 = 80 V across the 10 ohm (heater) load.
Using a modern high end high impedance digital voltmeter should give you 0.1 V resolution on a range reading 120 V. One can not realistically put a continuous 1500 W resistor for 120 V in this meter. You can do a lot of useful troubleshooting or circuit tracing with a meter with 0.1 V resolution, and an external 1500 W load.
.
One-eyed Jack
Senior Member
- Location
- Scotland Neck, NC
Two 100 watt bulbs in series make a very reliable neutral chaser. Cheap,readily available at your local hardware store.
AV ELECTRIC
Senior Member
I agree Ive done this before and saves a lot of time .
NewOnMyOwn
Member
- Location
- NJ
So even using a solenoid tester could leave room for error? I see, thanks guys.
brantmacga
Señor Member
- Location
- Georgia
- Occupation
- Former Child
I can't say I blame him. You gave the wrong diagnosis; so off that the customer went and purchased a new appliance when that was never the problem.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
I didn't suggest a dummy load since the OP has a perfectly good one: the dryer not working and the voltage imbalance showing up.
All he has to do is narrow down the point where the imbalance starts, geographically speaking, by testing at the accessible places.
I've seen people buy new dryers when the old one ran but didn't heat. Of course, the supplying only the heater element was open.
All he has to do is narrow down the point where the imbalance starts, geographically speaking, by testing at the accessible places.
I've seen people buy new dryers when the old one ran but didn't heat. Of course, the supplying only the heater element was open.
There's my answer.
To add, what I meant is that a low-impedance load (such as a solenoid tester, a 100w bulb, or a dryer's line-to-neutral loads) will expose the voltage differences, while a high-impedance meter won't.
The voltages must remain balanced (within a small range) under load for the load to function properly with line-to-neutral loads. Line-to-line-only loads have no 120v components.
Any ol' time.
There's my answer, again. Maybe i should read more and type less:roll:
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100207-1448 EST
zappy:
A typical digital voltmeter has a 10 megohm input resistance with a small amount of shunt capacitance.
A Simpson 260 or 270 on the 250 V range is 25,000*250 = 6.25 megohms.
One megohm = 1,000,000 ohms.
If the Wiggy was considered a pure resistance, which it is not, then 1000 ohms at 120 V would be 14,400/1000 = 14.4 W of power dissipation. That is a lot of power in a small space. Because the coil in the Wiggy has a substantial amount of inductance the dissipating resistance can be less than 1000 ohms while the circuit impedance could be 1000 ohms. My off hand guess is that if the AC impedance was reduced to 100 ohms there would be an excessive amount power dissipation.
I tried a Google search for a Wiggy impedance specification. SqD's page did not work. I did not find what I was looking for. But ran across the following that is sort of funny.
http://www.pumpsonline.com/electrical.htm
.
zappy:
A typical digital voltmeter has a 10 megohm input resistance with a small amount of shunt capacitance.
A Simpson 260 or 270 on the 250 V range is 25,000*250 = 6.25 megohms.
One megohm = 1,000,000 ohms.
If the Wiggy was considered a pure resistance, which it is not, then 1000 ohms at 120 V would be 14,400/1000 = 14.4 W of power dissipation. That is a lot of power in a small space. Because the coil in the Wiggy has a substantial amount of inductance the dissipating resistance can be less than 1000 ohms while the circuit impedance could be 1000 ohms. My off hand guess is that if the AC impedance was reduced to 100 ohms there would be an excessive amount power dissipation.
I tried a Google search for a Wiggy impedance specification. SqD's page did not work. I did not find what I was looking for. But ran across the following that is sort of funny.
http://www.pumpsonline.com/electrical.htm
.
- Location
- Illinois
- Occupation
- retired electrician
Gar,
My guess is that a wiggy draws about 30 mA at 120 volts. This is based on that fact that it sometimes trips a 30 mA ground fault breaker and other times it doesn't. I have never hooked a meter up to find out for sure.
My guess is that a wiggy draws about 30 mA at 120 volts. This is based on that fact that it sometimes trips a 30 mA ground fault breaker and other times it doesn't. I have never hooked a meter up to find out for sure.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100207-1718 EST
don:
Thanks.
That makes it about 4000 ohms impedance or thereabouts. Equivalent to a 3.5 W bulb load. However, the bulb maybe a better load for it has a lower resistance when cold.
This provides some useful values that may help many understand their instrument better.
.
don:
Thanks.
That makes it about 4000 ohms impedance or thereabouts. Equivalent to a 3.5 W bulb load. However, the bulb maybe a better load for it has a lower resistance when cold.
This provides some useful values that may help many understand their instrument better.
.
Ohmy
Senior Member
- Location
- Atlanta, GA
This is how we do it. Works like a charm. The old wiggies are great too but sometimes that do not put enough load on the circuit.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
The latter.
mivey
Senior Member
Isn't that for $100k+ talent?
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
"I'm so confused!" ~ Vinnie Barbarino
- Status