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I bought a power adapter to my printer but the lable on it says output voltage is 32V at 1.56A that means it is going to deliver around 50 Watts to my printer. On the other hand the original adapter used to supply 20 Watts at 32V and 0.625A. my question is would the new adapter, supplying 50 Watts, break down my printer? Thank you

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    \$\begingroup\$ As long as the voltage is correct and the power supply is capable of providing the maximum current required by the printer, then the power supply will be fine. Your printer will only take the power it needs. \$\endgroup\$ – HandyHowie Mar 8 '18 at 15:38
  • \$\begingroup\$ But what about the rest, I mean if my printer would the 20 Watts it needs, what would happen to the remaining 30 watts? Are they gonna dissipate into heat? But that means things will start melting down \$\endgroup\$ – Raykh Mar 8 '18 at 15:42
  • \$\begingroup\$ It's not going to supply the power. It's a maximum rating. \$\endgroup\$ – pjc50 Mar 8 '18 at 15:45
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    \$\begingroup\$ If you have a 700hp car and you drive like an old lady and only use 100hp... what happens to the extra 600hp? \$\endgroup\$ – Trevor_G Mar 8 '18 at 15:50
  • \$\begingroup\$ Ok good analogy. But the adapter is constantly supplying power just like you blocking the flow of high current river will overflow and flood everywhere. In case of electricity that means heat and component melt down \$\endgroup\$ – Raykh Mar 8 '18 at 15:54
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For replacing an adapter four conditions are important:

  • The voltage of the adapters should be the same (or within spec range), mostly around 10%.
  • The current that can be delivered by the adapter should be AT LEAST the current from the original adapter. The device will use current and gets it from the adapter. If the adapter is not capable of delivering it, than problems occur (not working, smoke etc).
  • The polarity should match (+ and - on the inner/outer pin). Some devices have a protection for this, some do not. So do not take the risk.
  • The size (length/diameter) of the plug should be same (trivial), otherwise it does not fit.

The wattage is just the multiplication of the current and voltage. Mostly there is a safety margin, but even than with the above conditions you should be ok.

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    \$\begingroup\$ +1 And the same polarity / AC or DC. \$\endgroup\$ – Trevor_G Mar 8 '18 at 15:44
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    \$\begingroup\$ @Trevor_G Thanks for this mandatory condition, I added it. \$\endgroup\$ – Michel Keijzers Mar 8 '18 at 15:46

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