I'm stumped.

[zappy]

If you go to www.autoshop101.com/forms/elec11.pdf and go to page 16. I don't understand the reading on the DMM. It shows .04 volts. So how could downstream then read 8.9 volts, If only .04 volts are getting thru? Could someone please explain what these readings mean. I need to brush up on my theory. Thank you.

 

[480sparky]

The connection has a small amount of resistance, therefore is a load in series. It will show a voltage across it that is related to the resistance.

 

[petersonra]

The situation is not well described.

Ideally, the voltage drop across a connector would be zero, just as it is across a conductor. However, in the real world, neither connectors nor conductors have zero resistance so there will be some voltage drop across them.

 

[zappy]

So that .04 volts is voltage drop? So the source voltage reads 12.7 volts. So why does downstream read 8.9 volts? Is 8.9 volts acceptable? 12.7 volts - 8.9 volts equals 3.8 volts dropped? Is that 3.8 VD from the .04 volt connection, or a combination of connections and wire resistance?

 

[nakulak]

first, start with ohms law v=ir, voltage = current x resistance

this is true at any point in the circuit for a resistance r

voltages in a series ckt add, so the voltage at the source = the sum of the series resistances


you need to brush up on the properties of series and parallel circuits

 

[gar]

100122-1552 EST

ZAPPY:

In the circuit you referenced there are many junction points, different wires, switches, and loads.

The source voltage at the points close to the actual source, ground some place on the chassis and the point at the top close to the power source, is 12.7 V. If the engine was off and you were measuring across the battery terminals, then the voltage might be slightly higher than 12.7 V.

Fundamentally the drawing shows a lot of series connected wires going to two bulbs in parallel.

A note at the bottom says there is little or no voltage drop from the ground side of the bulbs to the chassis. Thus you have a low resistance path in this part of the circuit relative to the load current.

As a frame of reference I will guess that the pair of lamps at normal voltage draw about 5 A. That would make each bulb about 12*2.5 = 30 W. This is a resistance per bulb of 4.8 ohms and in parallel 2.4 ohms. I will make an invalid assumption for simplicity. This is that the bulb resistance will be 4.8 ohms when the voltage across it is 8.9 V. This is not correct but still I will use 2.4 ohms load at 8.9 V. This means that the load current on the battery from this circuit is 8.9/2.4 = 3.7 A.

The difference in voltage between the source and the lamp load is 12.7 - 8.9 = 3.8 V. Thus the series resistance from the source to the top of the lamps is 3.8/3.7 = 1.03 ohms. The voltage drop across the one junction connector being measure is small compared to the 3.8 V drop. This junction connector is not the cause of the problem. So you might probe voltages at various points along the top path relative to the top end of the lamps.

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[petersonra]

So that .04 volts is voltage drop? So the source voltage reads 12.7 volts. So why does downstream read 8.9 volts? Is 8.9 volts acceptable? 12.7 volts - 8.9 volts equals 3.8 volts dropped? Is that 3.8 VD from the .04 volt connection, or a combination of connections and wire resistance?

It seems the point of the exercise is that somewhere in the circuit there is a voltage drop that should not be there because there is only 8.9 V across the lamps and it should be around 12V. That infers there is resistance in the circuit that should not be there. The extra resistance is NOT across the connector that reads 0.04V. It is somewhere else in the circuit.

A poorly done illustration and accompanying text because it is confusing to someone without a good practical understanding of Ohm's law.Not the best thing to be using for students who kind of by definition don't have such an understanding.

What I would do to debug this circuit would be to probe the circuit with the black lead of the VOM on the chassis and start at the lamp and work back toward the battery. The reading at the lamp is 8.9 volts to chassis. Its 12V to chassis at the start of the circuit. Somewhere in there it transitions from 12V to 8.9V and that is where the issue is.

 

[zappy]

Thank you everyone for your help.

Thank you everyone for your help.

I agree, they don't explain whats going on very well on that page.

 

[RETRAINDAILY]

So that .04 volts is voltage drop? So the source voltage reads 12.7 volts. So why does downstream read 8.9 volts? Is 8.9 volts acceptable? 12.7 volts - 8.9 volts equals 3.8 volts dropped? Is that 3.8 VD from the .04 volt connection, or a combination of connections and wire resistance?

no not good, so you check across the switch should be 0, check across connector, should be 0 but its not you got .4 thats the problum you have have high resistance. could be corroded blowing lamps. is what i get from reding page 15

 

[gar]

100123-0802 EST

RETRAINDAILY:

There is excessive voltage drop in the series path to the bulbs, but it is not in the junction connector with 0.04 V drop. Further high resistance in series with the lamps from a relatively constant voltage source, the car battery, will not cause bulb burn out, but will actually increase lamp life.

.

 

[K8MHZ]

Also, without being charged, the battery will drop voltage for any load applied. I would be more suspicious about the 12.7 volt reading from a battery supplying a load. To verify, connect your voltmeter to a car battery and read it. Then, with the car not running, turn on the headlights and watch the voltage drop right at the battery.

The voltage drop in 12 VDC is considerably more pronounced than what us 120 volt sparkys are used to seeing. In addition, there are many connectors used and each is an additional point of a drop in voltage.

 

[gar]

100214-1326 EST

Measurement from my car battery, 32 deg outside.

Load resistance 10 ohms.
No load voltage 12.56.
10 ohm loaded voltage 12.52.
Current = 1.25 A.
Small load change internal resistance of the battery = 0.04/1.25 = about 0.032 ohms.

At 100 A cranking current the internal resistance might be a little higher.

In the example referenced in post #1 it is reasonable classify the battery as a constant voltage source.

Maybe two to 4 times this 1.25 A current is the load from the circuit referenced.

Smaller batteries than mine will have somewhat higher internal resistance.

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[wallionmick]

Much needed Info..!! Thanks for sharing..!!

Wallion..!!


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