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Really basic question, why do we need to use power supplies in school experiments? Watch

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    Is it so we can limit the current, limit the voltage, or limit both?

    Also, how does a power supply reduce/increase the current/voltage to the level you want?

    Thanks
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    (Original post by blobbybill)
    Is it so we can limit the current, limit the voltage, or limit both?

    Also, how does a power supply reduce/increase the current/voltage to the level you want?

    Thanks
    Well even if you could directly use mains in a particular school experiment you wouldn't want to because it's at a hazardously high potential - so you want to prevent the possibility of someone handling a live conductor as far as possible, if you can make the same experiment at a lower volts you'd definitely prefer to for safety.

    Certainly some bench power supplies are voltage regulated (though iirc the ones we had in school physics weren't) - which makes them act more like an ideal battery with no internal resistance... which removes a confounding variable from your experimental design.

    https://www.youtube.com/watch?v=T7YcdOC5dZg has it correct... though one of the problems he doesn't mention with early switch mode psu's was that they created radio frequency interference which could jam radio signals (and also cause nearby computers to crash) though that's not really a problem with recent designs.
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    (Original post by Joinedup)
    Well even if you could directly use mains in a particular school experiment you wouldn't want to because it's at a hazardously high potential - so you want to prevent the possibility of someone handling a live conductor as far as possible, if you can make the same experiment at a lower volts you'd definitely prefer to for safety.

    Certainly some bench power supplies are voltage regulated (though iirc the ones we had in school physics weren't) - which makes them act more like an ideal battery with no internal resistance... which removes a confounding variable from your experimental design.

    https://www.youtube.com/watch?v=T7YcdOC5dZg has it correct... though one of the problems he doesn't mention with early switch mode psu's was that they created radio frequency interference which could jam radio signals (and also cause nearby computers to crash) though that's not really a problem with recent designs.
    Do the power supplies used for school experiments, do they reduce the voltage, current, or both the voltage and current?
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    (Original post by blobbybill)
    Is it so we can limit the current, limit the voltage, or limit both?

    Also, how does a power supply reduce/increase the current/voltage to the level you want?

    Thanks
    Yeah. Mains voltage combined with the available current/ power from the mains outlet is lethal.

    Bench supplies are used for safety and also because it's easier to design and build equipment to work at low d.c. voltages than direct from the mains a.c. supply.

    A typical bench power supply will use in order:

    1) Fusing and residual current/leakage detection to disconnect from the supply if too much current is taken or if a fault condition arises
    2) a transformer to isolate the mains input and reduce the a.c. voltage
    3) a diode rectifier and capacitors to produce a d.c. output
    4) some form of semiconductor voltage regulation to produce a controlled variable d.c. voltage output together with a variable output current limiter.
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    (Original post by uberteknik)
    Yeah. Mains voltage combined with the available current/ power from the mains outlet is lethal.

    Bench supplies are used for safety and also because it's easier to design and build equipment to work at low d.c. voltages than direct from the mains a.c. supply.

    A typical bench power supply will use in order:

    1) Fusing and residual current/leakage detection to disconnect from the supply if too much current is taken or if a fault condition arises
    2) a transformer to isolate the mains input and reduce the a.c. voltage
    3) a diode rectifier and capacitors to produce a d.c. output
    4) some form of semiconductor voltage regulation to produce a controlled variable d.c. voltage output together with a variable output current limiter.
    Ok thanks.. As well as reducing the voltage, do they reduce the current too?
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    (Original post by blobbybill)
    Ok thanks.. As well as reducing the voltage, do they reduce the current too?
    Think V = IR

    If the box sets the voltage to 6 V and the external resistance is 2 ohm, the current will be 3 A.

    The box will limit the current, in case the external resistance is low, e.g. a short circuit. Trip switches 4tw.
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    (Original post by Pigster)
    Think V = IR

    If the box sets the voltage to 6 V and the external resistance is 2 ohm, the current will be 3 A.

    The box will limit the current, in case the external resistance is low, e.g. a short circuit. Trip switches 4tw.
    So yes, they do reduce both voltage and current? Is that correct?? (yes or no)
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    Sometimes you need 'dc' as well
    These help provide a more 'dc' current than the AC from the mains as well
    :yep:

    (Original post by blobbybill)
    So yes, they do reduce both voltage and current? Is that correct?? (yes or no)
    You can control the amount of each, I believe.
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    (Original post by blobbybill)
    So yes, they do reduce both voltage and current? Is that correct?? (yes or no)
    There would be a safety limit on the current ; but the current through a circuit would depend on the draw of the devices on it.

    Typically you'd set the voltage on a bench PSU- say to 12v. If you put a bulb in line that might draw 1A. If you put 6 LEDs in line instead that might only draw 200mA at 12v
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    (Original post by Trinculo)
    There would be a safety limit on the current ; but the current through a circuit would depend on the draw of the devices on it.

    Typically you'd set the voltage on a bench PSU- say to 12v. If you put a bulb in line that might draw 1A. If you put 6 LEDs in line instead that might only draw 200mA at 12v
    Okay, but reducing the voltage from 12V to 1V in the same circuit would reduce the current? So the power supplies do reduce the current from the mains? Right?
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    (Original post by blobbybill)
    So yes, they do reduce both voltage and current? Is that correct?? (yes or no)
    Mains supply is 240 V. The current depends on the resistance. But if the current gets above a certain value, e.g. 13 A, then the fuse in the plug will burn out, or a trip switch will trip.

    Using a transformer, you can change the voltage to whatever value you need, which for schools will be up to about 12V - the sorts of values that cheap bulbs, diodes, thermistors etc will accommodate.

    The current will, once again, depend on the resistance. And once again, if it gets above a certain value will trip the trip switch built into the power supply.

    The box will reduce the voltage and will limit the current. It can't reduce the current as the value of the current isn't set by the box, but by the resistance of the external circuit.
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    (Original post by Pigster)
    Mains supply is 240 V. The current depends on the resistance. But if the current gets above a certain value, e.g. 13 A, then the fuse in the plug will burn out, or a trip switch will trip.

    Using a transformer, you can change the voltage to whatever value you need, which for schools will be up to about 12V - the sorts of values that cheap bulbs, diodes, thermistors etc will accommodate.

    The current will, once again, depend on the resistance. And once again, if it gets above a certain value will trip the trip switch built into the power supply.

    The box will reduce the voltage and will limit the current. It can't reduce the current as the value of the current isn't set by the box, but by the resistance of the external circuit.
    So the PSU would have the same resistance all the time? It has a fixed resistor?
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    P = I V

    If the output V = 12 V and I = 2 A, then P = 24 W.

    Assuming 100 % efficiency...

    Since the input V = 240 V, then the PSU will be drawing 0.1 A.

    Since the PSU is lowering V, it must be increasing I.
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    (Original post by Pigster)
    P = I V

    If the output V = 12 V and I = 2 A, then P = 24 W.

    Assuming 100 % efficiency...

    Since the input V = 240 V, then the PSU will be drawing 0.1 A.

    Since the PSU is lowering V, it must be increasing I.
    But that is different to ohms law, which says that when voltage increases, current increases proportionally, and vice versa with decreasing.

    How do you know whether you it is a Ohms law situation (where increasingvoltage also increases current), or whether it is a situation like you said (where P=VI, so increasing voltage decreases the current to achieve the same power).

    They are two different situations. In one, increasing voltage also increases current. In the other, increasing voltage decreases the current. Can you eplain that please?
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    The two are connected, P = I V and V = I R, therefore P = I2 R

    In my example, I said the output was 12 V and 2 A. The only way this is possible is if R = 6 ohm. If you decreased the external resistance, then the current would increase. This would then make the input current (to the PSU) to also increase.
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    (Original post by blobbybill)
    So the PSU would have the same resistance all the time? It has a fixed resistor?
    no it does not work like a resistor. ideally it's purely converting energy from electrical energy input to electrical energy output (but at a different potential) using a transformer as shown in the video I linked before.

    if nothing is connected to the PSU output there's no current output and no power output (because P=IV)
    the rule for transformers is
    IVout=IVin
    power out = power in

    so no power out means no power in... which means no current in (again because P=IV)
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    (Original post by Joinedup)
    no it does not work like a resistor. ideally it's purely converting energy from electrical energy input to electrical energy output (but at a different potential) using a transformer as shown in the video I linked before.

    if nothing is connected to the PSU output there's no current output and no power output (because P=IV)
    the rule for transformers is
    IVout=IVin
    power out = power in

    so no power out means no power in... which means no current in (again because P=IV)
    But it reduces the current and voltage from the mains to make it safer doesn't it?
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    (Original post by blobbybill)
    But it reduces the current and voltage from the mains to make it safer doesn't it?
    Well the current that goes through a person with a pd across them obeys the I=V/R law

    current is the killer so low voltage is better... skin resistance varies depending on how damp the persons skin is but about 12 V isn't considered sufficient to push a dangerous current through anyone however wet their hands are.
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    (Original post by Pigster)
    P = I V

    If the output V = 12 V and I = 2 A, then P = 24 W.

    Assuming 100 % efficiency...

    Since the input V = 240 V, then the PSU will be drawing 0.1 A.

    Since the PSU is lowering V, it must be increasing I.
    Okay. I get that the PSU is only drawing enough current alongside the 240V mains input to provide the power output desired.

    By "Since the PSU is lowering V, it must be increasing I", you mean in order to get the same power output, right?
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    (Original post by blobbybill)
    By "Since the PSU is lowering V, it must be increasing I", you mean in order to get the same power output, right?
    Assuming 100 % efficiency, the power that goes in, must come out therefore:

    Pin = Pout

    Iin x Vin = Iout x Vout

    If Vin is bigger than Vout, then Iin must be smaller than Iout
 
 
 
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