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difference in parameters when circuit switched on and when switched off

Discussion in 'General Electronics Discussion' started by krishna42099, Oct 1, 2012.

  1. krishna42099

    krishna42099

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    Oct 1, 2012
    Hi, I've got a weird doubt, say for example, I've got a new bulb written on its packaging as 12v and max current 0.2 amps, so I've connected it to a 12 volt suppy and added a resistor of 60 ohms (r=v/i, 12v/0.2 amps=60 ohms) in series. my question is generally in series voltage is shared on the components then if i connect this resistor in series to the bulb, so how does bulb gets its voltage from then???


    thx in advance for your time..
     
  2. duke37

    duke37

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    718
    Jan 9, 2011
    If you have two 60ohms reistors in series, then the voltage will be shared equally between them and the current will be 12/120 = 0.1A.

    If you have a resistor in series with a bulb, then things are different. The resistance of a bulb varies greatly with its temperature. If you measure the resistance of your 60ohm bulb with a meter, it will be much less than 60ohm. The voltage will be lower than expected across the bulb and more across the resistor.
     
  3. krishna42099

    krishna42099

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    0
    Oct 1, 2012
    thx duke37,..see you said at the last, the voltage will be more on the resistor, so is that remaining voltage on the bulb enough to light the bulb, another question, when I take the bulb out of circuit, if i checked voltage in between the resistor (which is connected to the positive) and the negative pole of the battery, I guess I still have 12 volts, but when I closed the circuit with the bulb in then I checked the voltage across the bulb, then it will not be 12 cos voltage shared between resistor and bulb, am I right??? can u pls explain me this...
     
  4. BobK

    BobK

    7,599
    1,641
    Jan 5, 2010
    A resistor drops the voltage only when current is flowing though it. This is explained by Ohm's law:

    E = I R

    When E is the voltage across the resistor, I is the current, and R is the resistance. If you make I = 0, you can see that E is zero, which is why your meter reads 12V when a resistor is placed between the battery and the meter, essentially no current is flowing. Actually, there is a small current flowing, and if you use a high enough resistance (try 1M Ohm) then you will see a voltage drop.

    Bob
     
  5. krishna42099

    krishna42099

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    0
    Oct 1, 2012
    thx bob...
     
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