Problems1 - 5.5 The circuit in Figure 5.6 is used to...

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Unformatted text preview: 5.5 The circuit in Figure 5.6 is used to rectify a 240-V RMS, 60-1-12 source. Load inductance is large. Required load power is 6 kW. Find: (a) average load current (b) peak load current (c) average diode current (d) peak diode reverse voltage. 5.9 In the circuit in Figure 5.11, the source is a 460V RMS, oO-HZ sinusoid. R = 20 Q, and the range of at is From 15° to 70°. The inductance is large. Find: (a) load average power for each extreme of a (b) SCR average current rating (c) SCR voltage rating. 5.10 The circuit in Figure 5.11 supplies 40 A to a load from a 230-V RMS, 60-1-12 source at or = 15°. Find: (a) average load power for or = 7’0O (b) SCR average current at or = 15°. 5.11 A Full-wave phase-control circuit such as that in Figure 5.11 supplies an average voltage ranging from 50 V to 200 V to a load of 5 Q. The source voltage is 2403/ RMS, 604-12. 5.12 In the circuit in Figure 5.13, V3 is a source of 4603.7 RMS, 604-12, and R is 20 £2. The value of or = 100°. Find: (at) average load current (b) SCR average current (c) FWD average current (d) peak SCR forward and reverse voltage. 5.22 In the circuit in Figure 5.31, the source voltage is ass—v RMS, (SO-Hz, and the load resistance is 15 9. For an output voltage of 300 V, find the minimum value of L needed to maintain continuous inductor Current. 5.23 in Problem 5.22,. the load voltage is variable from 200 V to 400 V. Find the minimum value of L needed to maintain continuous inductor current for all conditions. 5.25 'in a circuit such as that in Figure 5.31, the source voltage is 240—V RMS, oO—I-Iz. The average load power is required to be variable from 3 kW to 6 kW. The value of 06 may not be smaller than 10°. Inductor cirrrent is just continuous at one extreme value of or. Find: (a) the required range of or (b) the required value of L (c) the current rating of the bridge SCRs. 5.31 5.32 5.33 5.34 5.35 5.36 Repeat Problem 5.30 with the load changed to include a large induc- tance in series with the load resistance. The diodes in Figure 5.45 are replaced by SCRs to control the power to a resistive load. The source and the load are the same as those in Problem 5.30. The value of or is 700. Find: (a) average load voltage (13) peak load current (c) average load current (d) average SCR current (e) peak SCR forward voltage (f) peak SCR reverse voltage. The circuit in Figure 5.45 has the diodes replaced by SCRs and a load with a large inductance component; VS : 240-V RMS, 60-Hz; R = 20 Q; the range of or is 2.00 to 75°. Find: (a) the range of average load voltage (1)) the range of average load power (c) the SCR current rating required. In the circuit in Figure 5.45, the source is 460»V RMS, can; and the load power varies from 3 kW to 1 kW. The diodes are replaced by SCRs. The load has a large inductance component. The phase-control angle cannot be less than 10°. Find: (a) the required range of or 03) peak SCR current ((2) average SCR current ((1) peak SCR forward or reverse voltage. In the circuit in Figure 5.5.3, the source is aeo-v RMS, 60-1-12, and R is 2.5 Q. There is no load inductance. Find: (a) average load voltage (b) average load Current (c) average load power (d) average diode current (e) peak diode reverse voltage. In Figure 5.53, the load has an added inductor so that the load current does not vary with time- V = 230 V RMS ' , S - , 60-1-12, and R z 20 9. F d the RMS value of the line current. In ...
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This note was uploaded on 10/05/2011 for the course EECS 101 taught by Professor Hero during the Spring '11 term at National Taipei University.

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