Calculation of ripple voltage after rectifier (equation from TAOE)
Iload is the current drawn by the load.  They are making the simplifying assumption that the ripple voltage is small compared to the average voltage, so the load current is reasonably constant.

For example, let's say you have a full wave bridge driven from the secondary of a transformer that puts out 12 V RMS.  Since the power line voltage is sinusoidal, the peaks will be sqrt(2) higher than the RMS, so 17.0 V.  Figure 700 mV drop across each diode.  There are always two diodes in series with the output of a full wave bridge at any time, so that means the bridge drops 1.4 V.  The resulting peaks are then 15.6 V.

Now let's say you use a "large" filter cap and the load is a 100 &Omega; resistor.  If the cap were infinitely large, the DC would always be at 15.6 V and the load would draw (15.6 V)/(100 &Omega;) = 156 mA.

Their point is that the load current won't be much different as long as the filter cap is large enough so that the dips in voltage between the line cycle peaks are small compared to the 15.6 V.  Suppose the voltage actually dips 500 mV between peaks.  What would the load current be then?  At the bottom of the dip it will be (15.1 V)/(100 &Omega;) = 151 mA.  That's not so different from the no-dip figure of 156 mA.  And, that's the minimum.  The average will be roughly half way between these two, so 153.5 mA, although that level of precision in stating the numbers is silly.  Nonetheless, it shows that approximating the load current as constant in this case only introduces 1.6% error.  That's inconsequential when using 20% capacitors.

As a exercise, let's see roughly what size cap gives you the 500 mVpp ripple in this example. 

&nbsp; &nbsp; (156 mA)(8.33 ms)/(500 mV) = 2.6 mF

Note that this equation includes the same simplifying assumption Horiwitz and Hill used.  It assumes that the load is drawing 156 mA the whole time.  As we saw, it actually dips to 151 mA at the bottom of the ripples.  But think about what that actually matters.  This way you are told to use a slightly larger cap.  These things are usually 20% at best.  You're going to leave some margin and round up to the nearest common value anyway.  You'll need to use at least a 3.3 mF cap just due to tolerance.  In practise you'd probably use 3.6 mF cap, or decide you can live with a little more ripple.