Question

ng 3. a) Consider the circuit shown in Fig. 1 below, where the diode is assumed to be ideal. Find the average and RMS values of the output voltage, vo(t). A tua Fig.! VE) = A sin(wt) b) Consider the circuit shown in Fig. 2 below, where the diode is assumed to be ideal. Using your answer to part (a) above and basic properties of inductors in a switching circuit, write down the average value of the output voltage, Vo(t). 1 m - 3 st) Fig. 2 titt) - A Sint) c) Consider the circuit shown in Fig. 3 below, where the diode is assumed to be ideal. Using your answer to part (a) above and basic properties of capacitors in a switching circuit, write down the average current through the resistor, R2. (2) In De La 27100 Fig 3 Elt) = A sinet

Answer 1

Answer:

Given that:

Consider the circuit given below , where the diode is assumed to be ideal.

a)

@VID Asinut RO T OS

Ksource voltage

• The above circuit diagram represents single phase half wave diode rectifier with R load with respective wave from the output voltage and current are shown.
• During positive half cycle diode is forward biased , it therfore conducts wt = 0⁰ to wt = $\pi$ , During the positive half cycle output voltage V₀ = source voltage V_s and load current is i₀ = V₀ /R
• After wt = $\pi$ , diode is reverse biased , it therefore turned off and goes into blocking ratio

The averege value of output (or) load voltage

Vo = 1/27 | Asinut d(wt)

:: A=Vm

V = Vn/27 * 2

  
Vo=Vm)
​​​​​​ or
Vo = AA

RMS value of output voltage   

Vor = [1/27 v;sin’ut dut)1/2

Van = Vom/241/* - cosuit dwave

Vor = Vm 2

Similarly

1. = V/R = VmTR and lor = VoR = Vm 2R

b)

P-1

।। था । आT ot आ- भी आर

• The above circuit represents single phase halfwave diode rectifier with 'RL' load .
• Because the inductor is connected in series with the load voltage exists till under forward bias conditions

The average value of output voltage

  

Vo = 1/27 Vnsinwt dwt

V = Vm 271-cosB + 1

Vo = Vm 271 - cos

c)

Yi to tie, Vier $2,232 & R = 100 ilt=Asinut A=IM ict-Imsinwil

Assuming node voltage V.

By apply KCL at node 'V'

i(t) = i +iR1 +iR

i(t) = 678 +V/R1+V/Rx

Asinwt = 0 +V/R1 +V/R2

iR2 = Asinut - C4Y - V/R1 dt

iR, = Asinut - GV +V/RC

iR2 = Asinwt - Ve RC

iR2 = Asinwt - Vet

Asinwt = i(t) – Ver