A bench supply is built with a full-wave bridge on a 50 Hz mains transformer…

Step 1. The ripple frequency. $f_r = 2f = 100$ Hz.

Step 2. The capacitor. We want $C$, so $C = \dfrac{I}{f_r\,\Delta V} = \dfrac{0.12\ \text{A}}{100\ \text{Hz} \times 0.50\ \text{V}} = 2.4 \times 10^{-3}\ \text{F} = 2400\ \mu\text{F}$.

Step 3. The mean output. $V_{dc} = V_p - \dfrac{\Delta V}{2} = 15\ \text{V} - 0.25\ \text{V} = 14.75$ V.

Step 4. The ripple factor. $r = \dfrac{\Delta V}{2\sqrt{3}\, V_{dc}} = \dfrac{0.50\ \text{V}}{2\sqrt{3} \times 14.75\ \text{V}} = 0.00979$, which is $0.979\ \%$.

Step 5. Half the capacitance. $1200\ \mu$F gives $\Delta V = \dfrac{0.12\ \text{A}}{100\ \text{Hz} \times 1200 \times 10^{-6}\ \text{F}} = 1.00$ V.