Turn AC into DC: a schematic puzzle
Four diodes steer both halves of the wave the same way; one big capacitor fills the gaps.
Think of the bridge as four one-way doors around a square. When the top AC terminal is positive, current enters at the top-left corner, passes through the top diode to Vout, returns from the load to the bottom-left corner, and leaves through the bottom diode to the other AC terminal. Half a cycle later the AC reverses, and the other two diodes open instead: current still arrives at Vout from the same side. The load always sees the same polarity. Between peaks the output would fall to zero, so the 1000 µF capacitor acts as a reservoir: it fills at each peak and feeds the load through the gaps. At 100 mA and 60 Hz the ripple is roughly I ÷ (2 f C) ≈ 0.8 V, and the two diodes in the path cost about 1.4 V.
Diode 1N4007: Opens during the half-cycle when the left AC terminal is positive, letting current up into Vout. Its stripe, the cathode, points at Vout.
Diode 1N4007: Opens during the other half-cycle, when the right AC terminal is positive. Cathode up, toward Vout, so both halves of the wave arrive at the same terminal.
Diode 1N4007: The return door for the first half-cycle. Current that has done its work in the load comes back to the ground corner and leaves through here to the right AC terminal.
Diode 1N4007: The return door for the second half-cycle. Anode at the ground corner, cathode at the left AC corner: it only ever carries current back out to the transformer.
1000 µF smoothing capacitor: The reservoir. It tops up at every peak and carries the load through the valleys, so the output is a slightly rippled DC rather than 120 bumps per second. Polarised: plus to Vout, minus to ground.