Hey there! Sign in to join this conversationNew here? Join for free

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

    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Pigster)
    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
    Okay. but the Power supply would always have a Voltage in of 240 (in the UK)? Is that right?
    Online

    17
    ReputationRep:
    Yes.
    • Study Helper
    Offline

    21
    ReputationRep:
    (Original post by blobbybill)
    Okay. but the Power supply would always have a Voltage in of 240 (in the UK)? Is that right?
    In theory, yes.

    In practice it's a bit different since 240V is a quoted nominal supply voltage. The actual supply varies quite considerably dependent on the collective consumer power demand. (For instance at peak times when everyone rushes out during TV breaks to boil a kettle of water for tea, the supply voltage will drop. Which, incidentally, is how polls estimate a TV audience size!).

    The law states 230V +10%, -6% meaning it can (and does) vary between 216V and 253V
    • Thread Starter
    Offline

    8
    ReputationRep:
    (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.
    Using the equation P=VI

    How come in a circuit like we do in school experiments to measure the current against different resistors, both the current and voltage increase proportionally, whereas in something with P=VI, increasing the voltage also increased the current?

    I know what ohms law is, but how come with P=VI when you increase voltage, current decreases, and in ohms law, they both increase proportionally? Is P=IV only when there is a certain power output? If so, we used a lightbulb in our circuits and when we increased the voltage, the current also increased (ohms law), so why didn't the current decrease when the voltage increased, like it should do according to P=VI?
    Online

    17
    ReputationRep:
    (Original post by blobbybill)
    How come in a circuit like we do in school experiments to measure the current against different resistors, both the current and voltage increase proportionally, whereas in something with P=VI, increasing the voltage also increased the current?
    Assume (for ease of maths) that the bulb has a fixed resistance.

    According to V=IR, doubling the voltage will double the current. According to P=IV, doubling the voltage will directly double the P, but will double it again, since I was also doubled, i.e. 2xV -> 4xP. The bulb gets brighter.
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Pigster)
    Assume (for ease of maths) that the bulb has a fixed resistance.

    According to V=IR, doubling the voltage will double the current. According to P=IV, doubling the voltage will directly double the P, but will double it again, since I was also doubled, i.e. 2xV -> 4xP. The bulb gets brighter.
    Ah, because you double the voltage, you also double the current, so power is multiplied by 4? is that what you mean?
    Online

    17
    ReputationRep:
    Yes. It is a square law.

    Tripling V leads to a 9xP (assuming constant R)
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.