Let's have a household with 230 Volt AC supply. What happens with the current if I start to gradually activate more and more (purely) resistive loads, like electric heaters and ovens?

Because I=U/R, the current should be decreasing, as more resistance is added into the circuit (and the voltage is constant).

On the other hand, when a lot of heaters is switched on, a typical 10A circuit breaker will trip.

So what is causing the tripping of the circuit breaker in this situation? It should be overcurrent, but current can increase only if resistance is decreasing.

  • \$\begingroup\$ Resistors in parallel don't add together. But currents in parallel add together. \$\endgroup\$
    – markrages
    Sep 1, 2014 at 20:02

1 Answer 1


You actually add the loads in parallel. So R decreases. And the current increases.

$$I_{total} = V (\frac{1}{R1}+\frac{1}{R2}+...+\frac{1}{Rn})$$


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