PS13_Solution[2]

PS13_Solution[2] - Name Section I EE331 Homework PS13 —...

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Unformatted text preview: Name: Section: I; EE331 Homework PS13 — spring 2011 Assigned Problems: 0 Next two pages of this PS Additional Problems (Instructor Option): 0 Any as assigned by instructor Page 1 of 3 Name: Section: EE331 Homework PSl3 — spring 2011 Problem 1: Consider the circuit shown below V1 0 9 LEAD-ACID BATTERY GENERATOR Where VI = 25V, R1 = 0.059 , V2 = 28V, R2 = 0.049, V3 = 25V , and R3 = 0.029. Diodes D1 and D2 are ideal diodes. a. Determine the 4 possible circuit conﬁgurations (based on the states of D1 and D2) b. For the given data, establish the valid conﬁguration. 0. Find the battery current. Problem 2: Consider the following half-wave rectiﬁer where we desire to transmit 300W to a heater modeled by the 39 resistor. The diode is ideal. |heatl Vpksinﬂ t 3H a. Determine the required RMS load voltage. b. Find the required peak voltage of the source. c. Sketch one cycle of the load current ihea, , indicating its peak value as well. Page 2 of 3 Name: Section: EE331 Homework PSl3 — spring 2011 Problem 3: Let’s modify the circuit of Problem 2 to include a non-ideal transformer with an output resistance of 0.60 as shown below. The conducting diode will have a forward bias voltage drop of IV. D 0.6[I 3|] a. Determine the no-load peak voltage of the secondary given the same heater power requirement of Problem 2. b. If the primary is rated for 120Vrms, determine the required transformer turns ratio. 0. Sketch the primary-side current for rated conditions (use your calculated turns ratio). Problem 4: Consider the full-wave rectiﬁer circuit shown below where VS = 170V sin(377t) and Rm“, 21009 a. If the diodes are ideal, determine the power transmitted to the load resistance. b. If the conducting diodes have a 1V drop each, sketch one cycle of the load current. 0. If the diodes are ideal (no voltage drop), determine the ripple and average voltage across the load if we place a 150uF capacitor in parallel with R110“. Page 3 of 3 0‘ D; 0W 0‘4 OM 04‘? ON? 0"\ “\ ﬁ/ LBW/MA L » ‘362\ mth cm). a) WOMAN: (kW—w) if: v 0 ° 3 M pk (3.552) \jwtﬂ upan L (QW'JM + 2 L C3 § X =*r-+«J'(Q2Q: (guard /——\LW\W(2 ’(‘ﬁ’ \$01474qu 3 > 3Mol- ﬁNx e 341400453»: 4 '- OTMNJ LO Adoﬂ:\/ "v 9W: ‘%—l%SM(o.?ql)= VAVE : \lef 9i”: 90- Si?) : Q g ‘, ...
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PS13_Solution[2] - Name Section I EE331 Homework PS13 —...

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