If I plug a circuit into a wall socket (which runs at 120V) the circuit consists of a 240 ohm resistor, is the circuit only pulling .5 amps?
2007-03-07
10:00:25
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8 answers
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asked by
captinhankey
2
in
Science & Mathematics
➔ Engineering
If I plug a circuit into a wall socket (which runs at 120V) the circuit consists of a 240 ohm resistor, is the circuit only pulling .5 amps?
Also, does a step-down transformer only affect voltage, and not current? Please explain how it works.
2007-03-07
10:04:48 ·
update #1
For the first part of your question, yes, the current draw is 0.5 amps. Good old Ohms Law works just fine. The resistance divided into the voltage will give 0.5, which is the amount of current being drawn by the circuit.
For the transformer question, for all intents and purposes, without getting into the finer points of transformer operation, power in will equal power out. 120 volts on the primary of a transformer at 1 amp, os 120 watts of electrical power. The secondary will provide that amount of power. In a step down transformer, the voltage goes down, the current goes up. The deciding factor is the turns ratio. For a step down transformer, that ratio might be 10 to 1. For every 10 volts in the primary, only 1 volt is available in the secondary, but at 10 times the current of the primary. Power is basically conserved in that, as previously stated, power in equals power out., with out going into the higher specifics of transformer operation. There are other losses, but they can be ignored for general explanations. While you can reverse the connections for a step down transformer to make it a step up transformer, don't try it if the secondary voltage is less than the voltage you intend to apply to it. A transformer with a 12 volt secondary will be burned out if you apply 120 volts to it. The reason is that the amount of wire in the secondary will not support the higher voltage.
2007-03-07 20:01:34
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answer #1
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answered by Anonymous
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You have to remember the the 120V in the wall is AC, and it is actually 120V rms. Basically 120V is the average of the AC waveform. The peak of the waveform is closer to 170V. So at the peak of the waveform there is closer to .707 Amps running through the resistor. But on average the current is .5A. But for a design you need to account for the peak current.
For a transformer, you already know that the voltage gets divided by the turns ratio. The same thing works in reverse for the resistor. if you put the same 240 ohm resistor on the secondary of a 1:2 transformer. The voltage on the secondary is 60V rms (across the resistor). but to the primary, the resistor looks like it is 480 ohms (doubled in the way that the voltage was divided going the opposite direction. As a result the current needed becomes half.
2007-03-07 11:09:28
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answer #2
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answered by BrianW 3
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To answer the 2nd part.
The transformer voltage is proportional to the ratio of the windings. In an *ideal* transformer the power would constant; input and output, so the product of V * I would be the same on both sides. But there aren't any ideal transformers. The current is dependant on the size of the core, the wire size in the windings. For the proper transformer in the proper application a good approximation is: the current is inversely proportional to the winding ratio, with an efficiency factor (like 0.95).
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2007-03-07 10:17:04
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answer #3
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answered by tlbs101 7
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Ohm's regulation: V = I*R or E = I*R the place V (or E) is voltage in volts, I is present day in amperes, and R is resistance in Ohms. Voltage is largely electric potential in a circuit, subsequently can neither be created or destroyed (regulation of Conservation of potential). Voltage is needed in a circuit to create a potential distinction for the present to pass. present day is the pass of electrons in a closed circuit. The process the present relies upon on which theory you utilize. nicely-known present day says that present day flows out the helpful terminal of a potential grant and into the detrimental terminal. This, of direction, is fake on account that electrons pass from a extra detrimental area to a extra helpful area with the aid of a potential distinction. The pass of present day from the detrimental terminal to the helpful terminal is electron pass. Resistance is the contest of present day pass. All aspects in a circuit have some resistance (till they're superconductors) and for this reason dissipate a number of the great voltage in a circuit. the great present day continues to be a similar, from the initiating of the circuit to the top. those suggestions pass into Kirchoff's Voltage regulation and Kirchoff's present day regulation, so i'm going to stop there.
2016-11-23 14:04:30
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answer #4
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answered by ? 4
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The half amp is correct. The transformer changes the voltage and ohm's law still holds at the secondary.
2007-03-07 10:17:06
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answer #5
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answered by Gene 7
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E=IR
E=voltage
I=current in amps
R =resistance to current flow
2007-03-07 10:05:21
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answer #6
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answered by JOHNNIE B 7
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voltage is current carrying conductor ,
resistance unit is ohms
current is a flow of electron
2007-03-08 10:46:10
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answer #7
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answered by Anonymous
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V= IR....work it out..
2007-03-07 10:05:07
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answer #8
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answered by Anonymous
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