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Title: Converting Volts & Amperes to Watts - 36V 1400mA & 72V 700mA: Calculate Wattage

we3rty 45381 18
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How do I calculate wattage from 36 V 1400 mA and 72 V 700 mA?

36 V × 1.4 A = 50.4 W, and 72 V × 0.7 A = 50.4 W [#18012897][#18012927] Use P = U × I, and remember that 1000 mA = 1 A [#18012897] If you mean a power supply, its input power will be higher than its output power because some energy is lost as heat, so efficiency matters [#18013133][#18013476] The loss is calculated from the input power and efficiency; for example, at 90% efficiency, 56 W input means 5.6 W lost, and at 87% efficiency, 56 W input means 7.28 W lost [#18013476]
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  • #1 18012885
    we3rty
    Level 7  
    Posts: 54
    Rate: 31
    Hello,


    36 V and 1400 mA = how many Watt is it?

    72 V and 700 mA = how many Watt is it?

    Sorry for the strange question.
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  • #2 18012896
    niewolno2
    Level 40  
    Posts: 4689
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    36 X 1.4 how much is it?
    72 X 0.7 is how much is it?
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  • #3 18012897
    MACIEK_M
    Level 29  
    Posts: 1179
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    1V * 1A = 1W

    1000 mA = 1 A.

    and all clear. :D

    We are talking about direct current and resistive loads. :D
  • #4 18012902
    we3rty
    Level 7  
    Posts: 54
    Rate: 31
    Such an example was given by someone else, and I supposedly corrected it and replied that you should also write wrong first, and then I will give you how much this person wrote, how much according to her it is.
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  • #5 18012909
    niewolno2
    Level 40  
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    Then put your calculations here.
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  • #6 18012927
    we3rty
    Level 7  
    Posts: 54
    Rate: 31
    U * I = P

    So it is 50.4W in both cases?
  • #7 18012929
    niewolno2
    Level 40  
    Posts: 4689
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    And I don't want to be otherwise.
  • #8 18012946
    we3rty
    Level 7  
    Posts: 54
    Rate: 31
    How a miracle a person who knows electricity and probably electronics wrote that in both cases it is 56 W
    What would have to meet such a requirement to get 56W?
  • #9 18012974
    krzysiek_krm
    Level 40  
    Posts: 4612
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    we3rty wrote:
    What would have to meet such a requirement to get 56W?

    Describe this point in detail.
    Perhaps this "someone" took into account the efficiency of the power supply, for example, it would be 90%.
  • #10 18013052
    we3rty
    Level 7  
    Posts: 54
    Rate: 31
    It is about the use of LED in the home
    https://www.dom.pl/oswietlenie-led-zastosowanie-tasmy-led-w-nowczesnych-wnetrzach.html

    krzysiek_krm you're right, it's probably about the efficiency of the power supply

    How is the efficiency of a power supply per watt calculated?
    U * I = P
    36 * 1.4 = 50.4W
    72 * 0.7- 50.4W

    Is the calculations good here?
    How does the power supply come, then it looks different? How is it calculated?
  • #11 18013133
    kblg13
    Level 3  
    Posts: 72
    Rate: 22
    When you plug in the power supply, some power is emitted on it. The input power will always (at least a little) be greater than the output power, and the difference in input and output power results in wasted power on the power supply. Perhaps there was 56W at the input, and 5.6W was left on the power supply. Then the efficiency: 50.6W / 56W is actually 90%.
  • #12 18013181
    we3rty
    Level 7  
    Posts: 54
    Rate: 31
    How to find out that the input was 56W?
    How to find out if there is 5.6W left on the power supply?
    What kind of power supply would it have to be for it to come out or is it all given on the power supply?

    If the efficiency of the power supply is 87%, how much would it be in the end?
  • #13 18013476
    kacpo1
    Level 33  
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    we3rty wrote:
    How to find out that the input was 56W?

    The current consumption should be indicated on the rating plate.
    we3rty wrote:
    How to find out if there is 5.6W left on the power supply?

    Since the input power is 56W and the efficiency is 90%, the power supply will lose 56W x 10% = 5.6W.
    we3rty wrote:
    What kind of power supply would it have to be for it to come out or is it all given on the power supply?

    The efficiency depends on the design of the power supply and the load. It is impossible to give the model of the power supply based on the efficiency. With an efficiency of 90%, it is probably a switching power supply.
    we3rty wrote:
    If the efficiency of the power supply is 87%, how much would it be in the end?

    Input power 56W, efficiency 87%, so 56W x 13% = 7.28W and so much less would be at the output.
  • #14 18021810
    we3rty
    Level 7  
    Posts: 54
    Rate: 31
    Thanks for the clarification.

    Which would be better power supply:

    36V and 1400mA
    if
    72V and 700mA

    Is it better for the power supply to have a larger volt range and less amps or less volts more amps? Which would be more effective or efficient?
  • #15 18021889
    krzysiek_krm
    Level 40  
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    we3rty wrote:
    Thanks for the clarification.

    Which would be better power supply:

    36V and 1400mA
    if
    72V and 700mA

    Is it better for the power supply to have a larger volt range and less amps or less volts more amps? Which would be more effective or efficient?

    This is a wrong question.
    Before, you wrote
    we3rty wrote:
    It is about the use of LED in the home

    and the high-power LEDs are energized current , you need a power supply with a current efficiency equal to the current of light emitting diodes. The range of the output voltage must be such that your diodes "fit" there.
  • #16 18022003
    we3rty
    Level 7  
    Posts: 54
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    I am asking pure hypothetical.

    Let's assume it is a 70W LED

    Would the power supply be better in this case

    36V and 1400mA
    if
    72V and 700mA

    I would not like to fully power the LEDs only in about 50W, e.g. with the attached potentiometer in the power supply (20W LED would have a spare), so in this case, which power supply would be better?
  • #17 18022005
    kacpo1
    Level 33  
    Posts: 1698
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    Maybe it's better to start with reading about LED powering?
    Exact parameters are needed, such as diode forward voltage and diode current. The power supply must be a constant current one, the voltage range of which is within the diode voltage and the current of which corresponds to the diode current.
    we3rty wrote:
    e.g. with an attached potentiometer in the power supply

    With this LED power, the potentiometer is off.
  • #18 18022249
    krzysiek_krm
    Level 40  
    Posts: 4612
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    Rate: 598
    we3rty wrote:
    I am asking pure hypothetical.

    Let's assume it is a 70W LED

    Would the power supply be better in this case

    36V and 1400mA
    if
    72V and 700mA

    I would not like to fully power the LEDs only in about 50W, e.g. with the attached potentiometer in the power supply (20W LED would have a spare), so in this case, which power supply would be better?

    It is done completely differently.
    First, you choose the LED that interests you.
    You check in the documentation what parameters it has: current If and voltage Vf, usually a certain spread of this voltage is given for the current If, it results simply from the production characteristics spread.
    You buy yourself current LED power supply with the appropriate output current (Your If) and the appropriate output voltage range (must "cover" the Vf range of your diode, preferably with up and down reserve).
    If you want, you can buy a dimmable power supply.
    And that's it.
  • #19 18022259
    kj1
    Electrician specialist  
    Posts: 3274
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    Rate: 848
    The problem you raised cannot be explained in a few words

    we3rty wrote:
    Would the power supply be better in this case

    36V and 1400mA
    if
    72V and 700mA

    If the parameters of the power supply are given in this way, we are dealing with a voltage stabilizing power supply with a certain maximum current efficiency.
    If the current capacity is exceeded, the protection will trip and most likely disconnect the power.

    LED power supplies have the specified voltage range and current.
    e.g.
    Title: Converting Volts & Amperes to Watts - 36V 1400mA & 72V 700mA: Calculate Wattage

    Which of the power supplies listed by you will be the right one depends on how many, which and how connected LEDs the illuminator you want to power contains
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Topic summary

✨ The discussion revolves around calculating wattage from voltage and current values, specifically for 36V at 1400mA and 72V at 700mA. The formula used is U * I = P, leading to a calculated power of 50.4W for both cases. Participants also discuss the efficiency of power supplies, noting that input power may exceed output power due to losses, with examples of efficiency calculations provided. The conversation shifts to the suitability of different power supply specifications for LED applications, emphasizing the importance of matching voltage and current ratings to the LED requirements. The need for a constant current power supply is highlighted, along with considerations for dimmable options and the significance of LED specifications.
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FAQ

TL;DR: 1000 mA = 1 A, and “1V * 1A = 1W.” Use P = U × I: 36 V × 1.4 A and 72 V × 0.7 A both equal 50.4 W; with 90% efficiency, input must be higher. [Elektroda, MACIEK_M, post #18012897]

Why it matters: This FAQ helps DIYers and pros convert V/A to W correctly and pick safe, efficient LED drivers for home lighting.

Quick Facts

How many watts is 36 V at 1400 mA and 72 V at 700 mA?

Use P = U × I. Convert 1400 mA to 1.4 A and 700 mA to 0.7 A. 36 V × 1.4 A = 50.4 W. 72 V × 0.7 A = 50.4 W. Both cases have equal wattage. [Elektroda, MACIEK_M, post #18012897]

What formula converts volts and amps (or mA) to watts?

Power equals voltage times current: P = U × I. Convert milliamps to amps by dividing by 1000. Example: 12 V × 0.5 A = 6 W. “1V * 1A = 1W.” [Elektroda, MACIEK_M, post #18012897]

Someone told me both cases are 56 W. When can that be correct?

That can describe input power when efficiency is considered. If output is 50.4 W and efficiency is 90%, input is about 56 W. The 5.6 W difference is heat lost in the supply. Use Pin = Pout / η for this conversion. [Elektroda, kacpo1, post #18013476]

How do I compute input, output, and losses for a power supply?

Use efficiency η. Output equals input times η. Loss equals input times (1 − η). Example: 56 W input at 90% efficiency loses 5.6 W and delivers about 50.4 W. Those losses heat the power supply, not the load. [Elektroda, kacpo1, post #18013476]

How can I tell the input was 56 W or the loss was 5.6 W?

Check the rating plate for input consumption. If efficiency is listed, compute losses from it. With 56 W input and 90% efficiency, loss is 56 × 10% = 5.6 W. Output equals 56 W minus 5.6 W. [Elektroda, kacpo1, post #18013476]

Which is better for a 70 W LED if I only want about 50 W?

Pick by LED current and voltage, not by V vs A tradeoff. Choose a constant‑current LED driver matching the LED’s If. Ensure the driver’s voltage range covers the LED’s Vf. Use a dimmable driver to set power near 50 W. [Elektroda, krzysiek_krm, post #18022249]

Why is “36 V and 1400 mA vs 72 V and 700 mA” the wrong question for LEDs?

LEDs are current‑driven devices. “This is a wrong question.” Use a constant‑current driver that matches the LED current. Ensure the driver’s voltage range covers the LED’s forward‑voltage range. Efficiency depends on driver design, not V vs A alone. [Elektroda, krzysiek_krm, post #18021889]

Can I dim a high‑power LED with a potentiometer?

Avoid a series potentiometer at tens of watts. It wastes power and can overheat. Use a dimmable constant‑current driver or PWM control. “With this LED power, the potentiometer is off.” [Elektroda, kacpo1, post #18022005]

What happens if I power LEDs from a constant‑voltage supply?

A constant‑voltage supply regulates voltage and has a current limit. If your LED string exceeds that current, protection may trip and disconnect output. Proper LED drivers regulate current and list an allowable voltage range. [Elektroda, kj1, post #18022259]

How do I pick a constant‑current LED driver? (How‑To)

  1. Check your LED’s forward current (If) and forward voltage (Vf).
  2. Select a constant‑current driver with output current equal to If.
  3. Ensure the driver’s voltage range spans Vf with headroom; pick dimmable if you need brightness control. [Elektroda, krzysiek_krm, post #18022249]

What’s the difference between input watts and output watts on a supply?

Input power is always higher than output. The difference becomes heat in the supply. Example discussed: 56 W in versus about 50–51 W out, near 90% efficiency. That waste heat must be managed for reliability. [Elektroda, kblg13, post #18013133]

How do I convert mA to A quickly before calculating watts?

Divide by 1000. For example, 700 mA = 0.7 A and 1400 mA = 1.4 A. Then use P = U × I to get watts. This yields 50.4 W for both example cases. [Elektroda, MACIEK_M, post #18012897]
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