Full wave bridge rectifier load power

Circuit diagram

             D1            D2
   A o---+---|>|---+---|<|---+---o B
         |         |         |
         |        (+)        |
         |         |         |
         |     [RL = 5 Ω]    |
         |         |         |
         |        (−)        |
         |         |         |
         +---|<|---+---|>|---+
             D3            D4

   10 V ac input applied between A and B

Bridge: four diodes D1 to D4, with the ac input at A and B and RL between the (+) and (−) corners. During the half-cycle when A is positive, D1 and D4 conduct (A → D1 → RL → D4 → B). During the other half-cycle, D2 and D3 conduct (B → D2 → RL → D3 → A). In both half-cycles, the current flows through RL in the same direction, from (+) to (−).

Calculation

Assumption: the 10 V is the rms value of the ac input to the bridge. (If 10 V is the peak value, see the note below.)

1. Peak input voltage: Vm = 10 √2 = 14.14 V
2. Two diodes conduct in series in each half-cycle, so the drop is 2 × 0.7 = 1.4 V.

Peak load voltage: VL(p) = 14.14 − 1.4 = 12.74 V

3. For a full-wave rectified sine wave, the rms load voltage is VL(rms) = VL(p)/√2 = 12.74/1.414 = 9.01 V (Average voltage Vdc = 2VL(p)/π = 8.11 V)
4. Power dissipated in the load:

P = VL(rms)²/RL = (9.01)²/5 = 81.2/5 = 16.2 W

(Average power uses the rms value, not the dc value: using Vdc would give only 13.2 W, which is not the power dissipated.)

Peak load current = 12.74/5 = 2.55 A; each diode must be rated for at least 2.55 A peak and a PIV above 14.14 V.

Note: If 10 V is the peak ac voltage, VL(p) = 8.6 V, VL(rms) = 6.08 V, and P = 6.08²/5 = 7.4 W.

Answer: P ≈ 16.2 W (for 10 V rms input).