The Bridge Rectifier

Strings (SiPjAjk) = S7P5A51     Base Sequence = 12735     String Sequence = 12735 - 5 - 51

Expressions Of Pj Problems
The Bridge Rectifier
Math The half-wave rectifier converts only the positive half-cycle of an AC input signal. So, the negative half-cycle is unused. The bridge rectifier is a full-wave rectifier that coverts both cycles of the AC signal.

(a) Circuit 121.9 (a) is a bridge rectifier. Which diodes are conducting during the positive AC cycle and which diodes are conducting during the negative AC cycle?

The box of 121.9(b) and its AC input and DC output represents an Integrated Circuit (IC) Rectifier. An integrated circuit is a collection of electronic circuits on a silicon chip.
(b) Why are the resistor and capacitor added in the circuit of figure 121.9(b)?

(c) Suppose the bridge rectifier of circuit 121.9(a) is used to provide a 50 v, 5 A DC supply. Determine the resistance of the load that will draw exactly 5 A and the rms source voltage that realized the desired DC voltage. Assume an ideal diode.

The strings: S7P5A51 (Physical Change).

The math:
Pj Problem of Interest is of type change (physical change). The bridge rectifier is a change device. It changes each cycle of an AC signal to a DC signal. (a) D1 an D3 will conduct during the positive AC half-cycle. D2 and D4 will conduct during the negative AC cycle.

(b) The resistor, capacitor (RC) act as a low pass filter. The role of the filter is to remove most of the remaining AC component in the DC output. Rectifier circuits tend to pulsate about a mean average. The filter removes most of the fluntuations.

(c) Load resistance, RL = 50/5 = 10 ω

load voltage vL = RL (iL) = (RL /T)∫0 T i(t) dt
Where T = 2π/ω

So, vL = (RL /T)∫0 T/2 (vpeak/RL)sinωt dt

So, vL = 2vpeak/π = 2√2(vrms)/π = 50 V
So, vrms = 55.5 V. The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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