# Large capacitors “producing more I²R heating”?

On page 35 of The art of Electronics Third Edition, when describing the disadvantages of reducing ripple by choosing large capacitors, the author writes:

The very short interval of current flow during each cycle(only very near the top of the sinusoidal waveform) produces more I²R heating.

However, isn't Q = I²Rt? So a shorter interval should result in less heat - why is more heat produced?

• Think constant ripple current from a converter at some particular load which the capacitor a filtering. Time becomes irrelevant. – winny Jul 8 '16 at 6:00
• Because I is squared and t isn't, and their product I·t (or ∫Idt if you prefer) is constant for the same average current. – Dmitry Grigoryev Jul 8 '16 at 9:05
• Or, see it like this, high peak current in small bursts versus steady average current, same average value but much higher RMS current for the pulsed one. Irms^2*R losses. – winny Jul 8 '16 at 9:31

• @Andrew: You're right that W.s = J, but he's really saying $$P_{AVG} = \frac {4W \times 0.5s + 0 W \times 0.5s}{1s} = 2~W$$. All the seconds cancel out and you're left with watts. Bruce just left out the divisor. – Transistor Jul 8 '16 at 10:25