HELP!! Need answers for test!!!

"Jim" a écrit dans le message de news: snipped-for-privacy@g43g2000cwa.googlegroups.com...

with such question he can cheat all he want!!!!

this type of question are not even right..,missing lot of info

Reply to
petem
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Don't worry -- several of the answers Rojas gave were wrong. Just like in school, it's always risky to copy from someone else's answer sheet.

Reply to
Nomen Nescio

They're wrong. A resistor drops voltage and is a voltage limiting device. If you (for instance) use a variable resistor to control the speed of a fan (or light), the current remains the same regardless of the setting. The only thing you're doing is reducing the voltage to the load.

Reply to
Frank Olson

We drop the voltage and subsequently it reduces the current. Or we pass current through a resistor and it produces a voltage drop thereby limiting the current. Voltage and current through a resistor are dependent on each other. No current, no voltage drop. The only argument remaining is the language we choose to say the same thing. I think we are all saying the same thing in one way or another.

Bob

Reply to
Robertm

Yes, the current will drop because the voltage to the load has been reduced due to the voltage drop across the resistor. Voltage and current at the resistor are dependent on each other so I'd say it's a matter of perspective.

Bob

Reply to
Robertm

That's right...cheat at your own risk... :)

Jim Rojas

Reply to
Jim Rojas

We are also reducing the current if the resistance of the load remains fixed. A lamp and fan would be more difficult to calculate because impedance enters into fan calculations, and lamp resistance changes dependent on element temperature which is dependent on applied voltage and subsequent current. If we were to assume a resistive load of 12 ohms across a 12 volt supply, we have one amp flowing. Add a 12 ohm resistor in series with the load and we now have 6 volts across the limiting resistor and 6 volts across the load giving us a current of 500 ma through the load.

Bob

Reply to
Robertm

What??? Current in a series circuit stays the same. Halving the voltage through the load in the series circuit you describe above doesn't reduce the amount of current drawn from the source. Where did you go to skool??

Reply to
Frank Olson

Let's illustrate with some numbers. I=E/R and the current is the same in all parts of a series circuit. If we initially have only a 12 ohm load across a

12 volt supply, we have 12/12= 1 amp. If we then add a 12 ohm resistor in series with a 12 ohm load, we then have 24 ohms in the circuit and 12/24= 0.5 amps. You will drop 6 volts across the series resistor and 6 volts across the load. The current will be half of what it initially was, assuming a resistive load. That's what they taught me in school.

Bob

Reply to
Robertm

If we agree that it doesn't matter to any of us when we are in front of our respective BBQ cooking some amazingly better hamburger then the one bASS make,then we have a deal..

"Robertm" a écrit dans le message de news: dnfg3e$91ci$ snipped-for-privacy@news3.infoave.net...

Reply to
petem

Frank he is right that if you change the value of a the total resistance in circuit the current wont be the same...

don't you remember those formula?

I=I/R I mean current E mean Voltage R mean resistance..

so if I have a variable resistance ( a pot) is series with a fix load, lets say 100 ohms fixed load connected to pot that varied from 0 to 1000 ohms on a 10 volt supply

at minimum resistance of the pot the current will be 10/100 or 0.1 amp at maximum resistance of the pot the current will be 10/100+1000 or 10/1100 or 0.09 (rounded) amps so its about 1/10 of the max current..

the thing you describe about limiting power from a pot in a normal every day situation is not simple resistor in series of the load (light of fan) its in fact cutting parts of the ac phase of the load..so when on 50% of the time power is feed to the load the average power seen at the load is 50% lower then max but it still can be 110 volts across the load from time to time.( at 60 cycle a second this come fast) ;-)

"Frank Ols>> "Frank Olson" wrote in

Reply to
petem

They taught the same thing in my school. I saw the 1 amp load in your first example as being fixed (current in a series circuit is constant). You are correct (in all respects). Chalk it up to a "blonde moment"... :-)

Reply to
Frank Olson

At my age, when I get them, they are called senior moments. :-)

Bob

Reply to
Robertm

I refuse to admit I'm "old". The big "50" is coming up next year but I intend to ignore it. I'll hack into NIST and set the clock back to 1999 and enjoy a few more years of "40's". :-)

Reply to
Frank Olson

message

I had a feeling you weren't Y2K compliant Frank.....

Reply to
Jackcsg

Deny it all you want dino, you'll still be 50

Doug L

Reply to
Doug L

He doesn't realize that he's actually been 50 years old since his last birthday. Next birthday he'll start his 51'st year. ..........

Hey Frank ....... let me put it in perspective for you. ...........

50 years equals ....

Half a century! ......

How's THAT grab ya? (chuckle, chuckle)

Reply to
Jim

Arggghhhhhh!!!!!!

Reply to
Frank Olson

If it makes you feel any better I'm only 45

Reply to
Mark Leuck

Gator,

What is the test for?

BobbyD

Gator wrote:

Reply to
bdolph

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