Q 2) i have a 12v DC automobile head lamp rated 50W
a resistor is to be connected in series with the lamp to permit to operate on 24v. what is the resistor and power rating. i did mind like R=E^2/R R=12^2/50=2.88R. then power rating on the resistor is 50W. im i on the right track


Q 2) a circuit contains 8 incandescent lamps connected in series across 480 volt and one is defective and you got find which one. Meters available are voltmeter, ohmmeter and ammeter.
which meter would you use in shortest time determine defective lamp, how'd s it and why

i'd use ammeter cause got a low resistor and must be connected across circuit with load to limit flow of current. is that correct? i'm kinda not really sure--second opinion.

 

How would you use the ammeter to actually find the defective bulb? What would happen if you used the other meters?

 

What is the current around the circuit when one of the lamps has failed - presumably the bad lamp is open-circuit?

Where would you connect an ammeter to get a reading? Would this be harmful in any way?

 

first thing guys, is the first problem correct?

Beenther: i would connect the meter with the power supply. if used other meter like the ohmmeter the power must be off as for voltmeter already no how much voltage on the line

Adjuster: is an industrial plant and the bad lamp not open circuit cause th other are still on.

 

If the failed lamp is not open-circuit, so that the other lamps are on, why would the technician use a meter - do we assume that (s)he is BLIND!!!

 

i assume that though

 

What do you suppose might happen if you connected a voltmeter across each bulb with power on? What would happen if you used the ammeter?

You can still use the ohmmeter, but not as quickly as the other meters.

 

1st question: numerically correct , but you have made a typing slip on your first expression. P=...

2nd question: Using an ohm meter would be correct with the power off. That is the safest way, if the power can be shut down.

With the power on, you DO NOT know what voltage is on the line, at least not along all of the string of lamps.

The lamps are in SERIES. One of them is faulty. Normally an equal voltage is dropped across each lamp. What changes if a lamp fails, either short-circuit, or open-circuit?

Where would you connect an ammeter? If the faulty lamp is open, there is no current in the line before you connect the meter. Even if the lamp is still conducting, the current is the same along the whole string.

You would need to connect the ammeter so that it gave a different reading when you found the defective lamp. What would be the voltage on the other lamps if you did this.?

Would this be safe, or would you end up changing more than one bulb?

 

Depending how far apart the lamps are, possibly the fastest way needs a long return lead to the meter. Then you would not need to measure at every bulb.

How is this done? How many measurements would you make?

 

first question only values of V&W are given so it can't be P. what to no is what should the resistance and power rating of resistor?

Q,2)the voltage on each would be 60 cause total is 480. doing four measurement would be faster than doing all 8 since is in series at the loads.

 

first question only values of V@W are given so it can't be P. what to no is what should the resistance and power rating of resistor?

Q,2)the voltage on each would be 60 cause total is 480. doing four measurement would be faster than doing all 8 since is in series at the loads.

...i did mind like R=E^2/R R=12^2/50=2.88R. then power rating on the resistor is 50W....

You cannot have "R=E^2/R" This formula is for Power. I normally use P for power. Possibly you may use W, but not R.

R=12^2/50 is good though, and you have the right numbers, 2.88Ω, 50W.

Do you still think that an ammeter is the best choice, if the power is on when you have to do the tests? It is rather easy to have an accident with an ammeter - do you know why that is?

Doing four measurements would be indeed be faster, but I think you could do only three - do you know the half-split method?

 

not familiar with half split method yet would be sure to run that by the instructor on Monday.

with the power on using ammeter the too much power might blow it up. as for voltmeter which can be connected directly across a power source might do the job better

 

not familiar with half split method yet would be sure to run that by the instructor on Monday.

with the power on using ammeter the too much power might blow it up. as for voltmeter which can be connected directly across a power source might do the job better

Yes, that's the point I was hoping you would see. Provided the voltmeter and its probes and leads are correctly rated for the line voltage, you could connect the voltmeter across any two points in the circuit without expecting trouble.

Ammeters should be connected in series with a load. An ammeter would produce a short if connected across the supply. Results: - loud bang, spark, meter and / or other equipment damaged, blown fuses, and possible injury to the operator.

It might be OK to connect an ammeter briefly across each lamp in turn. Even so, when you connected across the failed lamp the remaining lamps would get a bit too much voltage.

The half-split method is worth learning: http://www.tpub.com/content/firetrucksandequipment/TM-9-254/css/TM-9-254_231.htm
This method is often described for tracing a "signal" through stages of an electronic system. In the case of the lamps, your "signal" is the voltage along the series chain of lamps. To use this method you would of course need a long lead to get your meter back to one end of the supply.

First test half way down the chain: decide if the fault is before that point or after it. If the fault is before, test half way up the first half of the chain. If the fault is after, test half-way down the second half of the chain. And so on...

This works very simply if the lamp is open - there will be no voltage past the failed lamp. If the lamp is shorted you can still use the method, but the test is harder. You would need to check if the voltage at each point was more or less than the 60V per lamp that would be expected with all the lamps good.

If you could disconnect the supply, you could also go half-splitting with an ohm-meter

 

BTW, you have learned a fair amount of basic troubleshooting methodology.