Master exam question

[ElToroLoco]

A coworker recently took his DC Masters exam and remembered one question that didn’t make sense. Can someone shed light on this question?

A series circuit has a 5 ohm resistor a 25 ohm lamp and a 5 ohm lamp. The voltage drop across the 5 ohm resistor is 10V

What is the source voltage?
A 50V
B 110V
C 220V
D 1110V

 

[LarryFine]

Welcome to the forum.

The answer I get is not one of the choices.

Spoiler

70v

 

[JoeStillman]

Welcome to the forum.

The answer I get is not one of the choices.

Spoiler

70v

Ditto.

 

[Dennis Alwon]

I don't know if this is correct but

Spoiler: Click Here

I saw 10v and 5 ohms which has a ratio 10/5 or 2 to 1. the rest of the resistors add to 30 ohms so 2 x 30 = 60v

60v and 10v = 70 volts. There might be an easier way, IDK

 

[mikeames]

I get the same as above.

Spoiler: My method

All three are in series so current is constant, and total VD = source V
5 ohms = 10v drop
25 ohms = 50v drop based on the known 10v drop
5 ohms = 10v

10+50+10=70v
R_T= 35
I_T = 2 amps

I_R1=2 amps
2 amps X 5 ohms = 10V
2 amps X25 ohms = 50v

 

[LarryFine]

Ohm's Law and the known values (volts and ohms) tell us the ratio that applies to the entire series circuit.

 

[Dennis Alwon]

Ohm's Law and the known values (volts and ohms) tell us the ratio that applies to the entire series circuit.

I drew it out because I have had no experience calculating resistors, etc. Once I drew it I saw a relationship and went from there. I did know that resistors in series are additive.

 

[TwoBlocked]

Show me a lamp that has the same resistance when there is no current and when there is. There is a reason light switches are rated with a "T" for tungsten.

 

[LarryFine]

Show me a lamp that has the same resistance when there is no current and when there is. There is a reason light switches are rated with a "T" for tungsten.

So, can you tell us the hot resistance? Otherwise, we have to use what's given?

 

[TwoBlocked]

So, can you tell us the hot resistance? Otherwise, we have to use what's given?

If you don't know the hot resistance, how could you know the voltage drop? In the real world, the question is BS. Likewise, when would you have two such lamps in series (they having the same current) resulting in one having 5 times the resistance, 5 times the wattage, and 5 times the voltage than the other? I dunno, maybe a plate protection circuit for a vacuum tube with the filament being one of the "lamps"? Been a while...

 

[LarryFine]

It was a rhetorical question to make my point. We have to presume that the voltage drop given applies to the load as being energized.

As I said, to answer a theoretical question, we have to use the info given and apply all-other-things (i.e., variables)-being-equal, or else the question cannot be answered.

I dunno, maybe a plate protection circuit for a vacuum tube with the filament being one of the "lamps"?

Or a 5-tube AM radio with the filaments in series? (50, 35, 12, 12, 12)

 

[TwoBlocked]

It was a rhetorical question to make my point. We have to presume that the voltage drop given applies to the load as being energized.

...

Yes, I agree, and the answer would be 70V. What gets me is, in the real world, HOW someone would know what the hot resistance of the lamps are. The only practical way would be to know the current through the circuit (either with an amp meter or using the voltage drop across the resistor, which would have little change under load) and then using the determined current and the measured voltage across each lamp to determine the hot resistance. Well, I suppose you could also determine the hot resistance if the characteristics of the lamps were known for the current. You would then design the circuit to have a specific voltage.

My point is, for the given resistances to be known, the voltages would have to first be known, which is what the question is asking. Maybe I'll think of an analogy.

 

[qcroanoke]

A coworker recently took his DC Masters exam and remembered one question that didn’t make sense. Can someone shed light on this question?

A series circuit has a 5 ohm resistor a 25 ohm lamp and a 5 ohm lamp. The voltage drop across the 5 ohm resistor is 10V

What is the source voltage?
A 50V
B 110V
C 220V
D 1110V

I'm sure glad stupid questions like that weren't on my masters test.
But I took mine with stone tablets and hammer and chisels were writing tools

 

[TwoBlocked]

I wonder which costs more. A 5 ohm lamp or a 25 ohm lamp.

 

[JoeStillman]

I wonder which costs more. A 5 ohm lamp or a 25 ohm lamp.

Definitely the 5 ohm. If power is inversely proportional to ohms, and cost is directly proportional to power, then lower ohms means higher cost.

 

[TwoBlocked]

Definitely the 5 ohm. If power is inversely proportional to ohms, and cost is directly proportional to power, then lower ohms means higher cost.

OK, then I'll go to my distributor and tell him to cancel the order for a 5 ohm lamp and substitute five 25 ohm lamps for the same price. Think he'll have a listing for lamps in those ohm ranges?

 

[JoeStillman]

OK, then I'll go to my distributor and tell him to cancel the order for a 5 ohm lamp and substitute five 25 ohm lamps for the same price. Think he'll have a listing for lamps in those ohm ranges?

I guess that would work if you connected them all in parallel, but would you come out ahead on cost? What's the price of ohms these days?

 

[ggunn]

I agree with posts 2-5. This is obviously a question meant to be a simple application of Ohm's Law; the variation in resistance with temperature of the lamp filaments is beyond the scope.

 

[TwoBlocked]

I ... What's the price of ohms these days?

Hahahaha

 

[ggunn]

What's the price of ohms these days?

I have heard that with inflation an ohm is now only worth 0.78V/I.