When we describe the working principle of the transformer, we say that we want to minimize the loss of electrical energy, therefore we reduce the current Q=I^2Rt, so heat reduces as well, but to keep the same power we increase the voltage, but Q=U^2/R *t. so why voltage doesn't give us the heat effect.
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Short answer: The power transferred depends on the voltage across the load, but the heat generated in the wire depends on the voltage between the two ends of the wire. The transformer increases the first one but reduces the second one (because the current through the wire is reduced). – The Photon Nov 30 '20 at 17:07
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that answer helped me can I accept that but wait for other answers as well? so if we have less current, we have less voltage drop U=IR, so less heat ? – math boy Nov 30 '20 at 17:08
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You can't accept it because it's a comment, not an answer. You can upvote the answer on the previous question (or on the even earlier question on the same topic linked from there). – The Photon Nov 30 '20 at 17:09
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So less current means less voltage drop on the wire using Ohm's law, so we have less heat? Am I correct? – math boy Nov 30 '20 at 17:10
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yes, correct.... – The Photon Nov 30 '20 at 17:10
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Q=U^2/R * t what U is in this formula ? – math boy Nov 30 '20 at 17:11
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U is the voltage across the resistor (wire in the case of the power transmission problem) that you want to work out how much heat it is generating. – The Photon Nov 30 '20 at 17:12
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but if we increase U, we get more heat, don't we? – math boy Nov 30 '20 at 17:14
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When you use the transformer, you increase $U_l$ (the voltage across the load), not $U_w$ (the voltage across the wire). The $U$ in the formula for heat generated by the wire is $U_w$, not $U_l$. – The Photon Nov 30 '20 at 17:17
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understood everything, thank you very much <3 – math boy Nov 30 '20 at 17:19