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I'm having serious trouble understanding this. If a component burns out when too much current is flowing through it, then how does turning on all stove-tops to their full setting burn out the oven's wall-plug?

Shouldn't turning on all the stove-tops increase resistance as you're adding more components to the circuit? Can someone explain (in simple terms) why this happens, also referring to mains supply and wall plugs and series/parallel phenomena. A diagram would help too I guess.

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  • \$\begingroup\$ Your question title seems incorrect. There is nothing in your post that suggests that your oven has burnt out or failed in any way. \$\endgroup\$
    – Transistor
    Commented Feb 22, 2020 at 11:17
  • \$\begingroup\$ Let me change it. \$\endgroup\$
    – El Flea
    Commented Feb 22, 2020 at 11:42
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    \$\begingroup\$ More resistances in parallel REDUCE the overall resistance. \$\endgroup\$
    – user16324
    Commented Feb 22, 2020 at 14:24

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If a fuse burns out when too much current is flowing through it, then how does turning on all knobs to their full setting burn out the wall-plug?

Turning on more rings will draw more current (in the same way that turning on more water taps will draw more water current). If the wall-plug burns out then it is inadequate for the job.

Shouldn't turning on all knobs increase resistance as you're adding more components to the circuit?

No. Turning on more cooker rings adds more resistors in parallel and that decreases reseistance.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Switching on additional heaters adds resistance in parallel. This reduces the overall resistance allowing more current to flow from the mains supply.

To understand how the power setting knobs work see my answer to Setting heat on electric stove.

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    \$\begingroup\$ You are one hell of a genius. Thank you, Sir Transistor! \$\endgroup\$
    – El Flea
    Commented Feb 22, 2020 at 11:47

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