Two factors may be working here. The first is the filament of the bulbs will have a different resistance with different temperatures (read different currents). So the resistance of the bulb will be variable.
The other is the battery's internal resistance. Depending on the amount of current being drawn, this could actually cause the terminal voltage of the battery to droop. Check this for each case.
2007-03-21 18:46:32
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answer #1
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answered by notBob 3
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Your question is a bit difficult to understand. If Im getting this correct, you have a single 6V battery that can either light 2 bulbs in series, or 2 bulbs in parallel?
If you have 2 bulbs in series with each bulb having an internal resistance of 4 ohms, then the total resistance of the circuit would be 8 ohms. With a 6V supply, the total current in the circuit would be .75 Amps. You should also measure a voltage drop across each individual bulb of approximately 3 volts. Each bulb in this circuit should be consuming roughly 2.25 Watts of energy.
In parallel, the total resistance of the circuit is going to be approximately 2 ohms. At 6V supply, the total current in the circuit would be roughly 3 Amps, with approximately 1.5 amps of current flowing through each bulb. The Power consumed by each bulb should be roughly 9 Watts.
2007-03-21 19:06:39
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answer #2
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answered by Jim M 2
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If you have two identical bulbs of resistance R, the total resistance when they are connected in series is 2R, and the equivalent resistance when they are connected in parallel is R/2. When you measure the voltage across each bulb when connected in series, you should get about 3 V, and when you measure the voltage across each bulb when connected in parallel you should get about 6 V.
The current draw of the parallel circuit will be about 4 times the current draw of the series circuit. Your voltage loss for the series connection appears to be 0.1 V and your voltage loss for the parallel appears to be about 0.5 V. this is about what it should be, although it indicates a pretty high resistance hookup.
BTW, V = IR. Dividing V by I will give you ohms, not amps.
2007-03-21 19:11:45
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answer #3
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answered by Helmut 7
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You will drop 6V across each circuit. Across 8 ohm (4+4) E/R=I is .75A With the 2 in parallel Req=R/n so 4/2= 2ohm eq. so E/R 6/2 = 3Amps Given DC circuits everything should be linear unless you got a Capacitor in there you didn't mention.
You are dropping 6V across the 2 bulbs in series so they each get 3V
You are dropping 6V across each of the 2 in parallel so I(current) is figured on the equivalent resistance.
Been a while but this is just a basic resistance circuit in both cases. Good luck.
2007-03-21 19:03:43
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answer #4
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answered by Anonymous
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Simple case: series: 4 ohms + 4 ohms = 8 ohms
6 volts/8 ohms = 0.750 or 750 mA.
parallel 4 ohms / 2 = 2 ohms
6 volts /2 ohms = 3 Amps
2007-03-21 19:33:04
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answer #5
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answered by Dusty 7
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Ohm
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This article is about the SI derived unit. For other meanings, see Ohm (disambiguation).
A multimeter can be used to measure resistance in ohms
A multimeter can be used to measure resistance in ohms
Several resistors. Their resistance, in ohms, is marked using a color code.
Several resistors. Their resistance, in ohms, is marked using a color code.
The ohm (symbol: Ω) is the SI unit of electrical impedance or, in the direct current case, electrical resistance, named after Georg Ohm.
Contents
[hide]
* 1 Definition
* 2 Explanation
* 3 Conversions
* 4 See also
* 5 External links
[edit] Definition
An ohm is a resistance in a conductor that produces a potential difference of one volt when a current of one ampere is flowing through it .
\Omega = \dfrac{\mbox{V}}{\mbox{A}} = \dfrac{\mbox{m}^2 \cdot \mbox{kg}}{\mbox{s}^{3} \cdot \mbox{A}^2}
[edit] Explanation
R is 1 ohm if V = one volt and I = 1 ampere
R is 1 ohm if V = one volt and I = 1 ampere
By definition from Ohm's Law, a device has a resistance of one ohm if a voltage of one volt causes a current of one ampere to flow (R = V/I). Alternatively and equivalently, a device that dissipates one watt of power with one ampere of current flowing through it has a resistance of one ohm (R = P / I 2).
Since 1990, the ohm has been maintained internationally using the quantum Hall effect, where a conventional value is used for the 'von-Klitzing constant', fixed by the 18th General Conference on Weights and Measures as R{K-90} = 25812.807 Ω.
The complex quantity impedance is a generalisation of resistance. Its real part is resistance and its imaginary part is reactance. Impedance, resistance and reactance all have units of ohms.
The symbol for the ohm is the Greek capital letter omega (Ω). If the Greek letter cannot be used, the word ohm is used instead. The various guides for the use of the International System of Units do not explicitly forbid the elision of the final "o" of some SI prefixes, although there is nothing in them to suggest that it is allowable, either. As a result, one is just about as likely to see "kilohm", "kiloohm" and even "kilo-ohm", and the same holds true for hecto-, micro-, nano-, pico-, femto-, atto-, zepto-, and yocto-. The only other SI unit to suffer from this kind of orthographic uncertainty is the ampere. In the particular case of the ohm, one even sees the "a" prefixes lose that vowel: hence megohm and gigohm. Higher prefixes are rarely used with ohm. In the other direction, milliohms (or millohms) are seen where the resistance of cables, etc., are measured.
Units of ohms, kilohms (103 Ω) and megohms (106 Ω) are used in electronic design documentation. On schematic diagrams and parts lists kilohms are abbreviated "K" and megohms are abbreviated "M". Thus, 33 kilohms would be rendered as 33K, and 5.1 megohms would be 5.1M. Another commonly used convention is that the multiplier is used to replace the decimal point, so that 5.1 megohms can also be represented as 5M1. This convention is used because a decimal point can be difficult to see in small or cluttered print. Values less than 1K are rendered either (a) without any symbol, or (b) with an "R", following the number; so 680 ohms can be shown as 680 or 680R. Resistors are usually identified by a reference designator, R, and a cardinal number, e.g., R12.
[edit] Conversions
A measurement in ohms is the reciprocal of a measurement in siemens, the SI unit of electrical conductance. Note that 'siemens' is both singular and plural. The non-SI unit, the mho (simply put, ohm written backwards), is equivalent to siemens but is mostly obsolete and rarely used.
[edit] See also
* Ohm's law
* Resistor
* abohm
[edit] External links
* Scanned books of Georg Simon Ohm at the library of the University of Applied Sciences Nuremberg
* Official SI brochure
* NIST Special Publication 811
* History of the ohm at sizes.com
Retrieved from "http://en.wikipedia.org/wiki/Ohm"
2007-03-21 18:48:47
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answer #6
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answered by Anonymous
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countless of your questions incorporate the two typographical errrors or inherently fake preparation. ability is measured in volt-amperes (additionally prevalent as watts). Resistance is measured in volt-in keeping with-ampere (additionally prevalent as ohms). you purely can not have a resistance of ???? watts. As for something ... the only (in keeping with risk) valid question may be the fourth one - its answer is the coil-and-bearing assembly. It is sensible - all the different configurations would properly be plagued by using the parameter you're attempting to degree.
2016-12-15 06:02:08
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answer #7
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answered by ? 4
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