4.9.SOLVED PROBLEMS 1. Determine the peak and rms voltages on the secondary of a transformer connected across a bridge rectifier to provide a no load dc voltage of 9 V. If the secondary winding resistance is 3 Ω and dynamic resistance of each diode is 1 Ω, determine the dc output across a load resistance of 100 Ω and 1 K. Also determine the regulation. Solution: 2. A 220 V, 60 Hz voltage is applied to a centre tapped step-down transformer of 22: 1 with a load of 1 K connected across the output of two-diode full-wave rectifier. Assume diodes to be ideal. If the resistance of half-secondary winding is 0.5 Ω, determine the (a) peak, rms and dc voltages, (b) peak, rms and dc currents, (c) dc power delivered to the load, (d) VA rating of the transformer secondary, (e) ac input to transformer assuming it to be 80% efficient, (f) ac ripple voltage across the load and its frequency, (g) How much is the PIVof each diode if the circuit is changed to a bridge rectifier using the full secondary winding of the same transformer? (h) How much are the peak, rms, dc voltages? Solution: (a) Vm (between full secondary) = 10√2 = 14.14 V vrms (full secondary) = 10 V, Vm (half secondary) = 7.07 V (f) As the ripple voltage under consideration would be the second harmonics because the magnitude of higher harmonics will become much less than it.(c) DC power delivered the load = I2DCRL = (4.2 mW (d) VA rating of the transformer can be found out using TUF. the ac ripple voltage is expressed as . Hence.495 mA)21 K = 20. ac-input power supplied.29(b) Solution: 4. and ripple factor. 4.29 (a).3. Also calculate the dc output power developed. A sinusoidal voltage of 22 V. 50 Hz is applied to a half-wave rectifier in Fig. Calculate the VDC.29(a)with dynamic resistance of the diode of 1 W. Calculate the maximum dc and rms currents flowing through the load.29(a) Figure 4. rectification efficiency. 4. Vr(rms). Figure 4. Figure 4. Solution: .29(b). IDC. Figure 4.29(a). Irms through a 1 KΩ load connected to a half-wave rectifier circuit shown in Fig. 4. Solution: Figure 4. rms currents that will flow through the load. Figure 4. dc and ac power developed across the load. ac voltages. Also calculate the dc.5.30.30.30 . A 230 V−0−230 V input voltage is connected to a full -wave rectifier shown in Fig. Calculate the dc. 4.31 Solution: Figure 4. Obtain number of turns in the primary and secondary of a transformer connected inFig. then it must have 10 turns in the secondary. Figure 4.31 to develop a dc voltage of 10 V across a load off 1 KΩ.Pin(ac) = V2rmsRL = 2302 × 1 K = 52. Calculate the ripple factor without and with capacitor filter in Fig. 7. Frequency of the ac voltage present across the load = fin = 50 Hz. Solution: . 4.32.32 If there are 100 turns in the primary.9 W 6.32. What is the frequency of the ac voltage present across the load? Figure 4.31. Figure 4. What is the frequency of ac voltage present across load? Figure 4. 4.33 Solution: .Theoretical ripple factor including filter = 8. Figure 4.33assuming all diodes to be ideal.33. What is the value of VDC and Vac developed across the load in circuit of Fig. Figure 4. Design a full-wave rectifier with LC filter to provide 10 V dc at 100 mA along with maximum ripple of 2%.34. Solution: 10. Figure 4.Frequency of the ac voltage present across the load = 2 fin = 100 Hz. The frequency of input voltage is 50 Hz. Figure 4. The value of the capacitor is 1000 μF and the value of load resistance is 100 Ω with frequency of input voltage equal to 50 Hz. 9.34 Solution: .34 is the output waveform of a half-wave rectifier with capacitor filter. Determine the ripple factor and dc voltage. Determine the ripple factor of the LC filter. Calculate the ripple factor in the case of a full-wave rectifier with π-filter having the component values C1 = C2 = 500 μF and laod resistance = 100 Ω. 12. Solution: .Ripple factor from theoretical expression is 11. Solution: Expression for ripple factor = r = Show that maximum dc power is transferred to the load in a full-wave rectifier only when the dynamic resistance of the diode is equal to the load resistance. Design a full-wave rectifier with an LC filter that can yield dc voltage of 9 V at 100 mA with a maximum ripple of 2%. (b) dc current through bleeder resistance.3 KΩ Figure 4. 1.e. C = 25 μF. RB = 700Lmax = 700 × 1. If the value of load resistance is RL connected in 10 KΩ. what would be the value of (a) filter dc output voltage. 13. Figure 4. then C = 46 μF = 50 μF (standard value). (c) dc current through the load. 4. Calculate the value of bleeder resistance required for the rectified voltage = 250sin 100πt.35 having the component values as L = 30 H. (d) ripple factor? Solution: Value of bleeder resistance (load resistance is open) = RB = 3ωL = 3 × 2πf × L = 28. A full-wave rectifier uses the LC filter as shown in Fig.3 H.13 H i. it will waste little power with the advantage of using L > LC.if LC is selected to be ten times higher than 0.35.3 H = 910Ω Since RB is ten times higher than the load resistance RL = 90 Ω.35 .