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

1. (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?
2. Yes.
3. (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
4. (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?
5. (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.
6. (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?
7. Yes. It is a square law.

Tripling V leads to a 9xP (assuming constant R)

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