# Confusion with unregulated PSUs: RMS current, transformer rating

I'm trying to understand a paragraph in Horowitz and Hill, "The Art of Electronics", under the sub-section "Transformers":

"Since a rectifier circuit draws current only over a small part of the cycle (during the time the capacitor is actually charging), the RMS current, and therefore the I^2R heating, is likely to exceed specifications for a load current approaching the rated RMS current of the transformer. The situation gets worse as you increase capacitor size to reduce pre-regulator ripple' this simply requires a transformer of larger rating."

What I do understand is that given a resistive load, for example, where ripple is small, current will only flow through the rectifiers for a small proportion of the cycle, therefore the current flowing in that small proportion will be a lot higher than that through the load at any time.

What I don't understand is why this would cause I2R problems with heating. Perhaps my understanding of RMS current is flawed. My contrary reasoning is that If for example 1 amp is drawn 90% of the time, the I2R heating would be the same as if 9 amps are drawn 10% of the time, since the heat dissipation is relatively slow compared to the 50Hz cycles.

I hope I've given enough information for someone to spot what I'm failing to understand.

Thanks

• If 1A is drawn across 1Ω for 0.9 second, 0.9 J is dissipated. If 9A is drawn across 1Ω for 0.1 second, 8.1 J is dissipated. Nov 8 '13 at 15:07